失眠网 > 【操作指导 | 代码实现】挑战程序设计竞赛2：算法和数据结构

【操作指导 | 代码实现】挑战程序设计竞赛2：算法和数据结构

时间：2020-08-31 23:59:57

书籍封面

第一章前言

1. 本人衷心建议

~~~~~~ 如果你是一位初学者，我指的是你只会基本的 C/C++ 编程，即使编的很烂，这本书对于你算法和数据结构的提升非常有帮助，所涉及的每一个算法案例都非常经典，且有非常细致的讲解，本人强烈推荐你理解好每一道题的细节。

~~~~~~ 如果你具备一定的算法与数据结构能力，本书的内容依然很适合你，你可以有针对性的补充相应章节的内容，一定会有意想不到的收获。

~~~~~~ 当你看到如此多的章节，每个章节还有如此多的内容时，一定不要被吓到，当本人对算法与数据结构一头雾水时，也是一步步算法过来的，这本书的学习大概花费了我近一个月的时间，所以你不必想着马上就可以看完很多，知识需要慢慢消化，Good luck ！

注：算法涉及的题目如何在网上搜索见下文，特意强调，无需翻墙。

2. 本书涉及的所有题目以及对应的编号

3. AOJ 使用方法《挑战程序设计竞赛》

网址

https://onlinejudge.u-aizu.ac.jp/home

注册

点右上角 Guest ==> Sign Up

注册完成后点击 submit

查找题目

1.点击网站上方 Challenge ==> Search

2.直接搜索题目

3. 或者根据上面 2 中题目对应的编号进行查找

点击 Volumes ，因为题号是 0202，前两位是 02，所以选 Volume2，然后找到对应编号即可

做题

进去之后选择语言，编写代码，并提交即可。当你看到全是日语的时候，不要惊慌，你完全可以看懂题，毕竟每道题的中文翻译都在本文中了…

提交后往下翻查看状态

用 virtual judge 做 AOJ 的题目

一个非常便捷的网站：/problem#OJId=Aizu&probNum=&title=&source=&category=all

首先进行账号注册

点进去做题就可以了

第二章算法与复杂度

2.4 算法的效率

2.4.1 复杂度的评估

2.4.2 大 O 表示法

2.4.3 复杂度的比较

第三章初等排序

3.1 挑战问题之前——排序

3.2 插入排序法

题解

代码实现

#include<iostream>#include<cstdio>#include<cmath>using namespace std;// 打印数列Avoid print(int A[],int n) {for (int i = 0; i < n; i++)printf("%d%c", A[i], i == n - 1 ? '\n' : ' ');}// 插入排序法void insertsort(int A[], int n) {// 从第二个开始,往前找比当前的小的值，// 比当前值大的就往后移，知道找到小于等于A[i]的for (int i = 1; i < n; i++) {int j = i - 1;int v = A[i];while (j >= 0 && A[j] > v) {A[j + 1] = A[j];j--;}A[j + 1] = v;print(A, n);}}int main() {int n, A[110];scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &A[i]);print(A, n);insertsort(A, n);return 0;}

3.3 冒泡排序法

题解

代码实现

#include<iostream>#include<cstdio>#include<cmath>using namespace std;// 冒泡排序法int bubbleSort(int A[], int n) {int cnt = 0;// 标记是否需要继续冒泡int flag = 1;for (int i = 0; flag; i++) {flag = 0;for (int j = n - 1; j >= i + 1; j--) {// 交换两个相邻的元素if (A[j] < A[j - 1]) {swap(A[j], A[j - 1]);flag = 1;cnt++;}}}return cnt;}int main() {int n, A[110];scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &A[i]);int cnt = bubbleSort(A, n);for (int i = 0; i < n; i++)printf("%d%c", A[i], i == n - 1 ? '\n' : ' ');printf("%d\n", cnt);return 0;}

3.4 选择排序法

题解

代码实现

#include<iostream>#include<cstdio>#include<cmath>using namespace std;// 选择排序法int selectionSort(int A[], int n) {int cnt = 0;for (int i = 0; i < n; i++) {int minj = i;for (int j = i; j < n; j++) {if (A[j] < A[minj])minj = j;}if (i != minj)swap(A[i], A[minj]), cnt++;}return cnt;}int main() {int n, A[110];scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &A[i]);int cnt = selectionSort(A, n);for (int i = 0; i < n; i++)printf("%d%c", A[i], i == n - 1 ? '\n' : ' ');printf("%d\n", cnt);return 0;}

3.6 希尔排序法

题解

代码实现

#include<iostream>#include<cstdio>#include<vector>using namespace std;//g数组vector<int>G;long long cnt;int A[1000000 + 10];// 间隔为g的插入排序void insertionSort(int A[], int n, int g) {for (int i = g; i < n; i++) {int v = A[i];int j = i - g;while (j >= 0 && A[j] > v) {A[j + g] = A[j];j -= g;cnt++;}A[j + g] = v;}}//希尔排序void shellsort(int A[], int n) {for (int i = G.size() - 1; i >= 0; i--) {insertionSort(A, n, G[i]);}}// 生成g数组void getArray(int n) {for (int h = 1; h <= n;) {G.push_back(h);h = 3 * h + 1;}}int main() {int n;scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &A[i]);getArray(n);shellsort(A,n);int len = G.size();printf("%d\n", len);for (int i = len - 1; i >= 0; i--)printf("%d%c", G[i], i == 0 ? '\n' : ' ');printf("%lld\n", cnt);for (int i = 0; i < n; i++)printf("%d\n", A[i]);return 0;}

第四章数据结构

4.1 挑战问题之前一一什么是数据结构

4.2 栈

题解

代码实现

stack1

/*实现栈的功能*/#include<iostream>#include<cstdio>using namespace std;// 实现栈的功能// top是指向栈顶的指针， s是实现栈结构的数组int top, S[1000];// 将x压入栈的操作void push(int x) {// top加1之后将元素插入到top所指的位置S[++top] = x;}// 将栈顶元素返回并删除int pop() {top--;return S[top + 1];}// 字符转数字int CharToInt(char s[]) {int ans = 0, i = 0;while (s[i] != '\0') {ans = ans * 10 + s[i] - '0';i++;}return ans;}int main() {char s[1000];int a, b;// 清空栈top = 0; while(scanf("%s",s)!=EOF){if (s[0] == '+') {b = pop();a = pop();push(a + b);}else if (s[0] == '-') {b = pop(); a = pop();push(a - b);}else if (s[0] == '*') {b = pop(); a = pop();push(a * b);}else {push(CharToInt(s));}}printf("%d\n",pop());return 0;}

stack2

/*利用STL的stack类*/#include<iostream>#include<cstdio>#include<stack>using namespace std;// 字符转数字int CharToInt(char s[]) {int ans = 0, i = 0;while (s[i] != '\0') {ans = ans * 10 + s[i] - '0';i++;}return ans;}int main() {char s[1000];int a, b;stack<int> S;while(scanf("%s",s)!=EOF){if (s[0] == '+') {b = S.top(); S.pop();a = S.top(); S.pop();S.push(a + b);}else if (s[0] == '-') {b = S.top(); S.pop();a = S.top(); S.pop();S.push(a - b);}else if (s[0] == '*') {b = S.top(); S.pop();a = S.top(); S.pop();S.push(a * b);}else {S.push(CharToInt(s));}}printf("%d\n", S.top());return 0;}

4.3 队列

题解

代码实现

queue1

/*实现队列功能*/#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int maxx = 100000 + 10;struct task {// 任务的名字和耗费的时间char name[100];int t;};task Queue[maxx];// 指向队头和队尾的指针int head, tail;// 判断循环队列是否已满bool isFull() {return head == (tail + 1) % maxx;}// 判断循环队列是否为空bool isEmpty() {return head == tail;}// 将元素x添加到队列的末尾void enqueue(task x) {Queue[tail] = x;// 循环队列tail = (tail + 1) % maxx;}// 返回队头元素并删除task dequeue() {task x = Queue[head];// 循环队列head = (head + 1) % maxx;return x;}int main() {int n, q;scanf("%d %d", &n, &q);for (int i = 0; i < n; i++) {scanf("%s %d", Queue[i].name, &Queue[i].t);}// 当前已经有n个元素head = 0; tail = n;// 当前时间int nowtime = 0;while (!isEmpty()) {// 取出队头task u = dequeue();// 执行时间片 p 和所需要时间u.tint c = min(q, u.t);// 任务执行了 c msu.t -= c;// 当前时间加上c msnowtime += c;// 任务还没做完，加入到队尾if (u.t > 0) {enqueue(u);}else {// 任务已经完成，打印出结果printf("%s %d\n", u.name, nowtime);}}return 0;}

queue2

/*利用STL的queue类*/#include<iostream>#include<cstdio>#include<algorithm>#include<queue>using namespace std;const int maxx = 100000 + 10;struct task {// 任务的名字和耗费的时间char name[100];int t;};queue<task> Queue;int main() {int n, q;scanf("%d %d", &n, &q);for (int i = 0; i < n; i++) {task t;scanf("%s %d", t.name, &t.t);Queue.push(t);}// 当前时间int nowtime = 0;while (!Queue.empty()) {// 取出队头task u = Queue.front();Queue.pop();// 执行时间片 p 和所需要时间u.tint c = min(q, u.t);// 任务执行了 c msu.t -= c;// 当前时间加上c msnowtime += c;// 任务还没做完，加入到队尾if (u.t > 0) {Queue.push(u);}else {// 任务已经完成，打印出结果printf("%s %d\n", u.name, nowtime);}}return 0;}

4.4 链表

题解

代码实现

List1

/*实现双向链表功能*/#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>using namespace std;const int maxx = 100000 + 10;struct Node {int key;Node *prev, *next;};Node *nil;// 链表初始化void init() {nil = (Node *)malloc(sizeof(Node));//C++ nil = new Node();nil->next = nil;nil->prev = nil;}// 从链表搜索出值为key的结点 O(n)Node* listSearch(int key) {// 从链表头结点开始访问Node *cur = nil->next;while (cur != nil && cur->key != key) {cur = cur->next;}return cur;}// 打印链表void printList() {Node *cur = nil->next;int flag = 0;while (cur != nil) {if (flag++ > 0)printf(" ");printf("%d", cur->key);cur = cur->next;}printf("\n");}// 删除结点tvoid deleteNode(Node *t) {// t为头结点不做处理if (t == nil)return;t->prev->next = t->next;t->next->prev = t->prev;free(t);//C++ delete(t);}// 删除头结点void deleteFirst() {deleteNode(nil->next);}// 删除尾结点void deleteLast() {deleteNode(nil->prev);}// 删除指定的key O(n)void deleteKey(int key) {deleteNode(listSearch(key));}// 在链表头部添加元素keyvoid insert(int key) {Node *x = (Node *)malloc(sizeof(Node));//C++ Node *x = new Node();x->key = key;x->next = nil->next;nil->next->prev = x;nil->next = x;x->prev = nil;}int main() {int n, key, size = 0;char op[20];scanf("%d", &n);getchar();// 初始化链表init();for (int i = 0; i < n; i++) {scanf("%s %d", op, &key);if (op[0] == 'i') {insert(key);size++;}else {if (strlen(op) > 6) {if (op[6] == 'F') {//删除头结点deleteFirst();}else {//删除尾结点deleteLast();}}else {//删除结点deleteKey(key);}size--;}}printList();return 0;}

List2

/*利用STL的list类*/#include<iostream>#include<cstdio>#include<cstring>#include<list>using namespace std;int main() {int n, key;char op[20];list<int> List;scanf("%d", &n);getchar();for (int i = 0; i < n; i++) {scanf("%s %d", op, &key);if (op[0] == 'i') {List.push_front(key);}else {if (strlen(op) > 6) {if (op[6] == 'F') {//删除头结点List.pop_front();}else {//删除尾结点List.pop_back();}}else {//删除结点for (list<int>::iterator it = List.begin(); it != List.end(); it++) {if (*it == key) {List.erase(it);break;}}}}}int flag = 0;for (list<int>::iterator it = List.begin(); it != List.end(); it++) {if (flag != 0)printf(" ");flag++;printf("%d", (*it));}printf("\n");return 0;}

4.5 标准库的数据结构

4.5.2 stack

4.5.3 queue

4.5.4 vector

4.5.5 list

第五章搜索

5.1 挑战问题之前一搜索

5.2 线性搜索

题解

代码实现

#include<iostream>#include<cstdio>using namespace std;// 线性搜索bool LinearSearch(int A[], int n, int key) {// 将key设置为数列最后一项A[n] = key;int i = 0;while (A[i] != key)i++;// 如果i不是n，则代表在原来的数列中找到key，// 否则代表原来数列中没有key，找到的是刚才插入到最后的元素return i != n;}int main() {int n, q, key;int sum = 0;int A[10000 + 10];scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &A[i]);scanf("%d", &q);for (int i = 0; i < q; i++) {scanf("%d", &key);if (LinearSearch(A, n, key))sum++;}printf("%d\n", sum);}

5.3 二分搜索

题解

代码实现

方式一

/*实现二分搜索*/#include<iostream>#include<cstdio>using namespace std;// 二分搜索bool BinarySearch(int A[], int n, int key) {int left = 0 , right = n, mid;while (left < right) {mid = (left + right) / 2;if (key == A[mid])return true;if(key > A[mid])left = mid + 1;if (key < A[mid])right = mid;}return false;}int main() {int n, q, key;int sum = 0;int A[100000 + 10];scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &A[i]);scanf("%d", &q);for (int i = 0; i < q; i++) {scanf("%d", &key);if (BinarySearch(A, n, key))sum++;}printf("%d\n", sum);}

方式二

/*利用STL的binary_search*/#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int main() {int n, q, key;int sum = 0;int A[100000 + 10];scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", &A[i]);scanf("%d", &q);for (int i = 0; i < q; i++) {scanf("%d", &key);if (binary_search(A, A + n, key))sum++;}printf("%d\n", sum);}

5.4 散列法

题解

代码实现

#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;const LL maxx = 1000000 + 10;char H[maxx][15];// 将字符转换为数值LL CharToInt(char ch) {if (ch == 'A') return 1;if (ch == 'C') return 2;if (ch == 'G') return 3;if (ch == 'T') return 4;return 0;}// 将字符串转成数值并生成keylong long getKey(char str[]) {LL sum = 0, p = 1;LL len = strlen(str);for (long long int i = 0; i < len; i++) {sum += p * CharToInt(str[i]);p *= 5;}return sum;}LL h1(LL key) {return key % maxx;}LL h2(LL key) {return 1 + (key % (maxx - 1));}// find 操作bool find(char str[]) {LL key, h;key = getKey(str);for (LL i = 0;; i++) {h = (h1(key) + i * h2(key)) % maxx;//找到keyif (strcmp(H[h], str) == 0) return true;// key不存在else if (strlen(H[h]) == 0) return false;}return false;}// insert操作void insert(char str[]) {LL key, h;key = getKey(str);for (LL i = 0;; i++) {h = (h1(key) + i * h2(key)) % maxx;//找到keyif (strcmp(H[h], str) == 0) return ;// key不存在else if (strlen(H[h]) == 0) {// 插入strcpy(H[h], str);return ;}}}int main() {int n, q, key;char str[20], op[20];for (int i = 0; i < maxx; i++)H[i][0] = '\0';scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%s %s", op, str);if (op[0] == 'i')insert(str);else {if (find(str))printf("yes\n");elseprintf("no\n");}}return 0;}

5.5 借助标准库搜索

5.5.1 迭代器

5.5.2 lower bound

5.6 搜索的应用——计算最优解

这是一个稍微有些难度的挑战题。如果各位觉得太难可以暂时跳过，等具备一定实力后再冋过头来挑战。

题解

代码实现

#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;LL A[100000 + 10];int n, k;//模拟装货过程bool check(LL P) {int cnt = k;for (int i = 0; i < n; ) {LL t = P;cnt--;// 卡车已经用完，证明卡车当前最大装载量不能装完货物if (cnt < 0)return false;while (i < n && t >= 0) {// 剩余重量可以装上当前货物if (t >= A[i])t -= A[i],i++;//装不下则用下一辆卡车去装if (t < A[i])break;}}return true;}// 二分找出P的最小值LL BinarySearch(LL sum) {LL left = 0, right = sum;LL mid;while (left + 1 < right) {mid = (left + right) / 2;//可以装好货物if (check(mid))right = mid;elseleft = mid;}return right;}int main() {LL sum = 0;scanf("%d %d", &n, &k);for (int i = 0; i < n; i++) {scanf("%lld", &A[i]);sum += A[i];}LL P = BinarySearch(sum);printf("%lld\n", P);return 0;}

第六章递归和分治法

6.1 挑战问题之前一递归与分治

6.2 穷举搜索

题解

代码实现

#include<iostream>#include<cstdio>using namespace std;int n, A[30];// 递归穷举bool solve(int i, int m) {// 已经找到得到m的组合方式if (m == 0)return true;// 已经穷举过A的全部元素了if (i >= n)return false;// 分别穷举不选A[i] 和选择A[i]的情况return solve(i + 1, m) || solve(i + 1, m - A[i]);}int main(){int t, M;scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &A[i]);}scanf("%d", &t);for (int i = 0; i < t; i++) {scanf("%d", &M);if (solve(0, M))printf("yes\n");elseprintf("no\n");}return 0;}

6.3 科赫曲线

题解

代码实现

#include<iostream>#include<cstdio>#include<cmath>using namespace std;const double PI = acos(-1.0);struct Point{double x,y;};void koch(int n, Point a, Point b) {// 递归结束if (n == 0)return;// 三等分点s,t ，(u 是 s,t 上方的点)Point s, t, u;// 角度从度转成弧度double th = PI * 60.0 / 180.0;s.x = (2.0 * a.x + 1.0 * b.x) / 3.0;s.y = (2.0 * a.y + 1.0 * b.y) / 3.0;t.x = (1.0 * a.x + 2.0 * b.x) / 3.0;t.y = (1.0 * a.y + 2.0 * b.y) / 3.0;u.x = (t.x - s.x) * cos(th) - (t.y - s.y) * sin(th) + s.x;u.y = (t.x - s.x) * sin(th) + (t.y - s.y) * cos(th) + s.y;koch(n - 1, a, s);printf("%.8lf %.8lf\n", s.x, s.y);koch(n - 1, s, u);printf("%.8lf %.8lf\n", u.x, u.y);koch(n - 1, u, t);printf("%.8lf %.8lf\n", t.x, t.y);koch(n - 1, t, b);}int main(){Point a, b;int n;scanf("%d", &n);// 初始化a.x = a.y = 0;b.x = 100, b.y = 0;printf("%.8lf %.8lf\n", a.x, a.y);koch(n, a, b);printf("%.8lf %.8lf\n", b.x, b.y);return 0;}

第八章树

8.2 有根树的表达

题解

代码实现

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;const int MAX = 100005;const int NIL = -1;struct Node {int p, l, r;};Node Tree[MAX];int n, Depth[MAX];void print(int u) {printf("node %d: ", u);printf("parent = %d, ", Tree[u].p);printf("depth = %d, ", Depth[u]);if (Tree[u].p == NIL)printf("root, ");else if (Tree[u].l == NIL)printf("leaf, ");elseprintf("internal node, ");printf("[");// 从子结点的最左边开始遍历int c = Tree[u].l;while (c != NIL) {printf("%d", c);if (Tree[c].r != NIL)printf(", ");// 跳到右侧紧邻的兄弟结点c = Tree[c].r;}printf("]\n");}// 递归得求深度void rec(int u, int depth) {Depth[u] = depth;if (Tree[u].r != NIL) // 右侧兄弟设置为相同深度rec(Tree[u].r, depth);if (Tree[u].l != NIL) // 最左侧子结点的深度设置为自己的深度 + 1rec(Tree[u].l, depth + 1);}int main() {int n;scanf("%d", &n);for (int i = 0; i < n; i++) {Tree[i].p = Tree[i].l = Tree[i].r = NIL;}int v, num, c, l, root;for (int i = 0; i < n; i++) {// v是当前结点值，num是v结点的子结点数 scanf("%d %d", &v, &num);for (int j = 0; j < num; j++) {scanf("%d", &c);//设置最左侧结点和紧邻右侧兄弟结点// l记录的是上一个结点，将当前的结点设置为l的紧邻右侧结点if (j == 0)Tree[v].l = c;elseTree[l].r = c;l = c;// 设置父结点Tree[c].p = v;}}// 找出根结点for (int i = 0; i < n; i++) {if (Tree[i].p == NIL)root = i;}rec(root, 0);for (int i = 0; i < n; i++)print(i);return 0;}

8.3 二叉树的表达

题解

代码实现

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;const int MAX = 100005;const int NIL = -1;struct Node {int parent, left, right;};Node Tree[MAX];int n, Depth[MAX], Height[MAX];int getDegree(int u) {int deg = 0;if (Tree[u].left != NIL)deg++;if (Tree[u].right != NIL)deg++;return deg;}void setDepth(int u, int depth) {if (u == NIL)return;Depth[u] = depth;setDepth(Tree[u].left, depth + 1);setDepth(Tree[u].right, depth + 1);}int setHeight(int u) {int h1 = 0, h2 = 0;if (Tree[u].left != NIL)h1 = setHeight(Tree[u].left) + 1;if (Tree[u].right != NIL)h2 = setHeight(Tree[u].right) + 1;return Height[u] = max(h1, h2);}int getSibling(int u) {if (Tree[u].parent == NIL)return NIL;//存在左兄弟if (Tree[Tree[u].parent].left != u && Tree[Tree[u].parent].left != NIL)return Tree[Tree[u].parent].left;//存在右兄弟if (Tree[Tree[u].parent].right != u && Tree[Tree[u].parent].right != NIL)return Tree[Tree[u].parent].right;return NIL;}void print(int u) {printf("node %d: parent = %d, sibling = %d, degree = %d, depth = %d, height = %d, ", u, Tree[u].parent, getSibling(u), getDegree(u), Depth[u], Height[u]);if (Tree[u].parent == NIL)printf("root\n");else if (Tree[u].left != NIL || Tree[u].right != NIL)printf("internal node\n");elseprintf("leaf\n");}int main() {int n;int u, l, r, root;scanf("%d", &n);for (int i = 0; i < n; i++) {Tree[i].parent = Tree[i].left = Tree[i].right = NIL;}for (int i = 0; i < n; i++) {scanf("%d %d %d", &u, &l, &r);Tree[u].left = l;Tree[u].right = r;Tree[l].parent = u;Tree[r].parent = u;}for (int i = 0; i < n; i++) {if (Tree[i].parent == NIL)root = i;}setHeight(root);setDepth(root, 0);for (int i = 0; i < n; i++)print(i);return 0;}

8.4 树的遍历

题解

前序遍历：F, B, A, D, C, E, G, I, H

中序遍历：A, B, C, D, E, F, G, H, I

后续遍历：A, C, E, D, B, H, I, G, F

层序遍历：F, B, G, A, D, I, C, E, H

代码实现

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;const int MAX = 100005;const int NIL = -1;struct Node {int parent, left, right;};Node Tree[MAX];int n;//前序排序void PreorderTreeWalk(int u) {if (u == NIL)return;printf(" %d", u);PreorderTreeWalk(Tree[u].left);PreorderTreeWalk(Tree[u].right);}//中序排序void InorderTreeWalk(int u) {if (u == NIL)return;InorderTreeWalk(Tree[u].left);printf(" %d", u);InorderTreeWalk(Tree[u].right);}//后序排序void PostorderTreeWalk(int u) {if (u == NIL)return;PostorderTreeWalk(Tree[u].left);PostorderTreeWalk(Tree[u].right);printf(" %d", u);}int main() {int n;int u, l, r, root;scanf("%d", &n);for (int i = 0; i < n; i++) {Tree[i].parent = Tree[i].left = Tree[i].right = NIL;}for (int i = 0; i < n; i++) {scanf("%d %d %d", &u, &l, &r);Tree[u].left = l;Tree[u].right = r;Tree[l].parent = u;Tree[r].parent = u;}for (int i = 0; i < n; i++) {if (Tree[i].parent == NIL)root = i;}printf("Preorder\n");PreorderTreeWalk(root);printf("\nInorder\n");InorderTreeWalk(root);printf("\nPostorder\n");PostorderTreeWalk(root);printf("\n");return 0;}

8.5 树遍历的应用 —— 树的重建

题解

图中后序遍历的结果为：3 5 6 4 2 8 9 7 1

代码实现

1、已知前序遍历和中序遍历，求后续遍历。

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<vector>using namespace std;vector<int> pre, in, post;int pos;int n, k;void reconstruction(int left, int right) {// 遍历到叶结点if (left >= right)return;// 当前子树的根int root = pre[pos++];// 找出当前根的值在中序数组里的下标int m = distance(in.begin(), find(in.begin(), in.end(), root));reconstruction(left, m);reconstruction(m + 1, right);//存入到后序数组post.push_back(root);}void solve() {pos = 0;reconstruction(0, pre.size());for (int i = 0; i < n; i++) {printf("%d%c", post[i], i == n-1?'\n':' ');}}int main() {scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &k);pre.push_back(k);}for (int i = 0; i < n; i++) {scanf("%d", &k);in.push_back(k);}solve();return 0;}

2、已知后续遍历和中序遍历求前序遍历

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<vector>using namespace std;vector<int> pre, in, post;int pos;int n, k;void reconstruction(int left, int right) {// 遍历到叶结点if (left >= right)return;// 当前子树的根int root = pre[pos--];// 找出当前根的值在中序数组里的下标int m = distance(in.begin(), find(in.begin(), in.end(), root));reconstruction(m + 1, right); // 注意优先建立左边的树，根节点时从后往前的reconstruction(left, m);//存入到后序数组post.push_back(root);}void solve() {pos = n - 1;reconstruction(0, pre.size());for (int i = n - 1; i >= 0; i--) {printf("%d%c", post[i], i == n-1?'\n':' ');}}int main() {scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &k);pre.push_back(k);}for (int i = 0; i < n; i++) {scanf("%d", &k);in.push_back(k);}solve();return 0;}

3、已知前序遍历和后序遍历不可求中序遍历

4、已知前序||中序||后序遍历（仅该树为完全二叉树），求层序遍历

#include<bits/stdc++.h>using namespace std;int n,a[100],tree[100];int idx=1;void f(int x){if(x > n) return ;tree[x] = a[idx++]; //已知前序遍历，求层序遍历f(x << 1);//tree[x] = a[idx++]; //已知中序遍历，求层序遍历f((x << 1) + 1);//tree[x] = a[idx++];//已知后序遍历，求层序遍历}int main(){scanf("%d",&n);for(int i = 1; i <= n; i++)scanf("%d",&a[i]);f(1);printf("层次遍历：");for(int i = 1; i <= n; i++)printf("%d ",tree[i]);return 0;}

5、已知层序遍历，求前中后遍历（仅满足树为完全二叉树时）

#include<stdio.h>const int N=100;int pre[N],in[N],post[N],tree[N],idx;;int n;void f(int x){pre[idx++]=tree[x]; //求前序遍历if(x<<1<=n)f(x<<1);//in[idx++]=tree[x]; //求中序遍历if((x<<1)+1<=n) //记得加括号，不然会先加1后乘2f((x<<1)+1);//post[idx++]=tree[x]; //求后序遍历}int main(){scanf("%d",&n);for(int i=1;i<=n;i++) //输入层遍历scanf("%d",&tree[i]);f(1);for(int i=0;i<n;i++) //输出printf("%d ",pre[i]);return 0;}

第九章二叉搜索树

9.2 二叉搜索树——插入

题解

代码实现

9.3 二叉搜索树——搜索

题解

代码实现

9.4 二叉搜索树——删除

题解

代码实现

#include<iostream>#include<cstdio>using namespace std;const int MAX = 100005;struct Node {int key;Node *parent, *left, *right;};Node *root, *NIL;//前序排序void PreorderTreeWalk(Node *u) {if (u == NIL)return;printf(" %d", u->key);PreorderTreeWalk(u->left);PreorderTreeWalk(u->right);}//中序排序void InorderTreeWalk(Node *u) {if (u == NIL)return;InorderTreeWalk(u->left);printf(" %d", u->key);InorderTreeWalk(u->right);}void insert(int value) {Node *y = NIL, *x = root;Node *z = new Node();z->key = value;z->left = z->right = NIL;// 从根结点往下遍历while (x != NIL) {y = x;// z 比 x 小，则从x的左侧遍历// z 比 x 大，则从x的右侧遍历if (z->key < x->key)x = x->left;elsex = x->right;}z->parent = y;// 没有父结点的是rootif (y == NIL) {root = z;}else {// z 比 y 小，则在y的左侧// z 比 y 大，则在y的右侧if (z->key < y->key)y->left = z;elsey->right = z;}}// 首先找到待删除值对应的节点，返回节点Node * find(Node *u, int value) {// 从根结点开始遍历下去while (u != NIL && u->key != value) {if (value < u->key)u = u->left;elseu = u->right;}return u;}// 树的最小值Node *treeMinimum(Node * x) {while (x->left != NIL)x = x->left;return x;}// case 3 ,搜索后一个结点Node *treeSuccessor(Node *x) {if (x->right != NIL)return treeMinimum(x->right);Node *y = x->parent;while (y != NIL && x == y->right) {x = y;x = y->parent;}return y;}// 传入的节点 z 即需要删除的节点void treeDelete(Node * z) {Node *y, *x; // 要删除的对象和 y 的子结点// 确定要删除的结点if (z->left == NIL || z->right == NIL)y = z;elsey = treeSuccessor(z);// 确定 y 的子结点xif (y->left != NIL)x = y->left;elsex = y->right;// 让子结点链接父结点if (x != NIL)x->parent = y->parent;// 让父结点链接子结点if (y->parent == NIL)root = x;else {if (y == y->parent->left)y->parent->left = x;elsey->parent->right = x;}// case 3if (y != z)z->key = y->key;delete y;}int main() {int n, value;Node * f;char s[200];scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%s", s);if (s[0] == 'i') {scanf("%d", &value);insert(value);}else if (s[0] == 'f') {scanf("%d", &value);f = find(root, value);if (f != NIL)printf("yes\n");elseprintf("no\n");}else if (s[0] == 'd') {scanf("%d", &value);treeDelete(find(root, value));}else {InorderTreeWalk(root);printf("\n");PreorderTreeWalk(root);printf("\n");}}return 0;}

第十章堆

10.2 完全二叉树

题解

代码实现

#include <iostream>#include <cstdio>using namespace std;// 分别求出结点i的父结点，左子结点、右子结点int parent(int i) {return i / 2; }int left(int i) {return 2 * i; }int right(int i) {return 2 * i + 1; }int main(){int H[300];int n;scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d", &H[i]);for (int i = 1; i <= n; i++) {printf("node %d: key = %d, ", i, H[i]);if (parent(i) >= 1)printf("parent key = %d, ", H[parent(i)]);if (left(i) <= n) printf("left key = %d, ", H[left(i)]);if (right(i) <= n)printf("right key = %d, ", H[right(i)]);printf("\n");}return 0;}

10.3 最大/最小堆

题解

代码实现

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxx = 2000000;int A[maxx + 1];int n;void maxHeapify(int i) {int left = 2 * i, right = 2 * i + 1;int largest;// 从左子结点、自身、右子结点选出值最大的结点，然后继续往下递归if (left <= n && A[left] > A[i])largest = left;elselargest = i;if (right <= n && A[right] > A[largest])largest = right;if (largest != i) {swap(A[i], A[largest]);maxHeapify(largest);}}int main(){scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d", &A[i]);for (int i = n / 2; i >= 1; i--)maxHeapify(i);for (int i = 1; i <= n; i++)printf(" %d", A[i]);printf("\n");return 0;}

10.4 优先级队列

题解

代码实现

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxx = 2000000;const int INF = 1 << 30;int A[maxx + 1];int n;void maxHeapify(int i) {int left = 2 * i, right = 2 * i + 1;int largest;// 从左子结点、自身、右子结点选出值最大的结点，然后继续往下递归if (left <= n && A[left] > A[i])largest = left;elselargest = i;if (right <= n && A[right] > A[largest])largest = right;if (largest != i) {swap(A[i], A[largest]);maxHeapify(largest);}}int extract() {int maxv;if (n < 1)return -INF;maxv = A[1];A[1] = A[n--];maxHeapify(1);return maxv;}int insert(int key) {n++;int i = n;A[n] = -INF;if (key < A[i])return -INF;A[i] = key;while (i > 1 && A[i / 2] < A[i]) {swap(A[i], A[i / 2]);i /= 2;}}int main(){char op[20];int k;while (scanf("%s", op) != EOF&&op[2] != 'd') {if (op[0] == 'i') {scanf("%d", &k);insert(k);}else {printf("%d\n", extract());}}return 0;}

10.5 通过标准库实现优先级队列

第十一章动态规划

11.1 动态规划法的概念

11.2 斐波那契数列

题解

代码实现

#include <iostream>#include <cstdio>using namespace std;int main(){int F[45],n;F[0] = F[1] = 1;for (int i = 2; i <= 44; i++)F[i] = F[i - 1] + F[i - 2];scanf("%d", &n);printf("%d\n", F[n]);return 0;

11.3 最长公共子序列

题解

代码实现

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxx = 1010;int c[maxx][maxx];int lcs(char s1[],char s2[]) {int m = strlen(s1), n = strlen(s2);for (int i = 1; i <= m; i++)c[i][0] = 0;for (int i = 1; i <= n; i++)c[0][i] = 0;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (s1[i] == s2[j])c[i+1][j+1] = c[i][j] + 1;elsec[i+1][j+1] = max(c[i][j+1], c[i + 1][j]);}}return c[m][n];}int main(){char s1[maxx], s2[maxx];int t;scanf("%d", &t);for (int i = 0; i < t; i++) {scanf("%s %s", s1, s2);printf("%d\n",lcs(s1,s2));}return 0;

11.4 矩阵链乘法

题解

代码实现

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxx = 110;const int INF = 1 << 30;int main(){int n, p[maxx], m[maxx][maxx];scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d %d", &p[i - 1], &p[i]);for (int i = 1; i <= n; i++)m[i][i] = 0;for (int l = 2; l <= n; l++) {for (int i = 1; i <= n - l + 1; i++) {int j = i + l - 1;m[i][j] = INF;for (int k = i; k <= j - 1; k++)m[i][j] = min(m[i][j], m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j]);}}printf("%d\n", m[1][n]);return 0;}

第十二章图

12.1 挑战问题之前——图

12.2 图的表示

题解

代码实现

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxx = 110;int main(){int n, m[maxx][maxx]; // 邻接矩阵int u, num, v;scanf("%d", &n);memset(m, 0, sizeof(m));for (int i = 1; i <= n; i++) {scanf("%d %d", &u, &num);for (int j = 1; j <= num; j++) {scanf("%d", &v);m[u][v] = 1; // u 与 v 之间画一条边}}for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)printf("%d%c", m[i][j], j == n ? '\n' : ' ');return 0;}

12.3 深度优先搜索

题解

代码实现

#include <iostream>#include <cstdio>#include <cstring>#include <stack>using namespace std;const int maxx = 110;int n, m[maxx][maxx];int u, num, v;int d[maxx], f[maxx];int flag[maxx];int nowtime = 1;void dfs(int u) {// 入栈时间flag[u] = 1;d[u] = nowtime++;for (int i = 1; i <= n; i++) {if (flag[i] == 0 && m[u][i] == 1) {dfs(i);}}//出栈时间f[u] = nowtime++;}int main(){scanf("%d", &n);memset(m, 0, sizeof(m));memset(f, 0, sizeof(f));for (int i = 1; i <= n; i++) {scanf("%d %d", &u, &num);for (int j = 1; j <= num; j++) {scanf("%d", &v);m[u][v] = 1;}}for (int i = 1; i <= n; i++) {if(flag[i] == 0)dfs(i);}for (int i = 1; i <= n; i++) {printf("%d %d %d\n", i, d[i], f[i]);}return 0;}

12.4 广度优先搜索

题解

代码实现

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int maxx = 110;int n, m[maxx][maxx];int u, num, v;int flag[maxx];int d[maxx];void bfs() {queue<int>q;q.push(1);d[1] = 0;while (!q.empty()) {int u = q.front();q.pop();for (int i = 2; i <= n; i++) {if (m[u][i] == 1 && flag[i] == 0) {// 标记为已经访问flag[i] = 1;// 更新长度d[i] = d[u] + 1;// 压入栈内q.push(i);}}}}int main(){scanf("%d", &n);memset(m, 0, sizeof(m));for (int i = 1; i <= n; i++) {scanf("%d %d", &u, &num);// 初始化flag[i] = 0;d[i] = -1;for (int j = 1; j <= num; j++) {scanf("%d", &v);m[u][v] = 1;}}bfs();for (int i = 1; i <= n; i++) {printf("%d %d\n", i, d[i]);}return 0;}

12.5 连通分量

题解

代码实现

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <vector>using namespace std;const int maxx = 100010;int n, m;int u, num, v;int flag[maxx];int color[maxx];vector<int> G[maxx];int c = 1;void dfs(int t) {stack<int>s;s.push(t);color[t] = c;while (!s.empty()) {int u = s.top(); s.pop();int len = G[u].size();for (int i = 0; i < len; i++) {int v = G[u][i];if (flag[v] == 0) {color[v] = c;flag[v] = 1;s.push(v);}}}}int main(){int q;memset(flag, 0, sizeof(flag));scanf("%d %d", &n, &m);for (int i = 0; i < m; i++) {scanf("%d %d", &u, &v);//将顶点v添加到G[u]数组中G[u].push_back(v);G[v].push_back(u);}for (int i = 0; i < n; i++) {if (flag[i] == 0) {dfs(i);c++;}}scanf("%d", &q);for (int i = 0; i < q; i++) {scanf("%d %d", &u, &v);if (color[u] == color[v]) {printf("yes\n");}else {printf("no\n");}}return 0;}

第十三章加权图

13.1 挑战问题之前——加权图

最小生成树

最短路径

13.2 最小生成树

题解

代码实现

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <vector>using namespace std;const int maxx = 1010;const int INF = 1 << 30;int n, m[maxx][maxx];int u, num, v;int flag[maxx];int prim() {// d数组是保存V-T， p数组是记录MST的int d[maxx], p[maxx];for (int i = 0; i < n; i++) {d[i] = INF;p[i] = -1;flag[i] = 0;}d[0] = 0;while (1) {int minv = INF;u = -1;// 比较出权值最小的边for (int i = 0; i < n; i++) {if (minv > d[i] && flag[i] == 0) {u = i;minv = d[i];}}if (u == -1)break;// 标记已经加入MSTflag[u] = 1;// 更新未访问的点与已经访问的点之间的最小的边（相对于每个点）for (int v = 0; v < n; v++) {if (flag[v] == 0 && m[u][v] != INF && d[v] > m[u][v]) {d[v] = m[u][v];p[v] = u;}}}int ans = 0;for (int i = 0; i < n; i++)if (p[i] != -1)ans += m[i][p[i]];return ans;}int main(){memset(flag, 0, sizeof(flag));scanf("%d", &n);for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {scanf("%d", &m[i][j]);if (m[i][j] == -1)m[i][j] = INF;}}printf("%d\n", prim());return 0;}

13.3 单源最短路径

题解

代码实现

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <vector>using namespace std;const int maxx = 1010;const int INF = 1 << 30;int n, m[maxx][maxx];int u, v ,k, c;int flag[maxx];void dijkstra() {int d[maxx];for (int i = 0; i < n; i++) {d[i] = INF;flag[i] = 0;}d[0] = 0;while (1) {int minv = INF;u = -1;// 比较出权值最小的边for (int i = 0; i < n; i++) {if (minv > d[i] && flag[i] == 0) {u = i;minv = d[i];}}if (u == -1)break;// 标记已经加入Sflag[u] = 1;// 更新经由u结点，V-S的v结点的最小成本for (int v = 0; v < n; v++) {if (flag[v] == 0 && m[u][v] != INF && d[v] > d[u] + m[u][v]) {d[v] = d[u] + m[u][v];}}}for (int i = 0; i < n; i++)printf("%d %d\n", i, d[i] == INF ? -1 : d[i]);}int main(){scanf("%d", &n);for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {m[i][j] = INF;}}for (int i = 0; i < n; i++) {scanf("%d %d", &u, &k);for (int j = 0; j < k; j++) {scanf("%d %d", &v, &c);m[u][v] = c;}}dijkstra();return 0;}

堆优化的单源最短路径

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <algorithm>using namespace std;const int maxx = 10010;const int INF = 1 << 30;int n;int flag[maxx];// 加权有向图的邻接表表示法，pair.first是顶点v，pair.second是权值vector<pair<int, int> > adj[maxx];void dijkstra() {priority_queue<pair<int, int> > PQ;int d[maxx];for (int i = 0; i < n; i++) {d[i] = INF;flag[i] = 0;}d[0] = 0;PQ.push(make_pair(0, 0));while (!PQ.empty()) {pair<int, int> f = PQ.top(); PQ.pop();int u = f.second;// 标记已经加入Sflag[u] = 1;//如果不是最小最短路径则忽略if (d[u] < f.first *(-1))continue;int len = adj[u].size();for (int j = 0; j < len; j++) {int v = adj[u][j].first;if (flag[v] == 0 && d[v] > d[u] + adj[u][j].second) {d[v] = d[u] + adj[u][j].second;// priority_queue默认取最大值，所以乘以-1PQ.push(make_pair(d[v] * (-1), v));}}}for (int i = 0; i < n; i++)printf("%d %d\n", i, d[i] == INF ? -1 : d[i]);}int main(){int u, v, k, c;scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d %d", &u, &k);for (int j = 0; j < k; j++) {scanf("%d %d", &v, &c);adj[u].push_back(make_pair(v, c));}}dijkstra();return 0;}

第十四章高等数据结构

14.1 互质的集合

题解

代码实现

#include <iostream>#include <cstdio>#include <vector>using namespace std;class DisjointSet {public :// rank记录树的高度vector<int> rank, p;DisjointSet() {};DisjointSet(int size) {rank.resize(size, 0);p.resize(size, 0);for (int i = 0; i < size; i++)makeSet(i);}void makeSet(int i) {p[i] = i;rank[i] = 0;}bool same(int x, int y) {return findSet(x) == findSet(y);}void unite(int x, int y) {link(findSet(x), findSet(y));}void link(int x, int y) {// 包含x的树的高度更高，则将y的树合并到x上if (rank[x] > rank[y])p[y] = x;else {p[x] = y;if (rank[x] == rank[y])rank[y]++;}}int findSet(int x) {if (x != p[x])p[x] = findSet(p[x]);return p[x];}};int main(){int n, a, b,t, q;scanf("%d %d", &n, &q);DisjointSet ds = DisjointSet(n);for (int i = 0; i < q; i++) {scanf("%d %d %d", &t, &a, &b);if (t == 0)ds.unite(a, b);elseprintf("%d\n", ds.same(a, b) ? 1 : 0);}return 0;}

第十五章高等图算法

15.1 所有点对间最短路径

题解

代码实现

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>using namespace std;typedef long long LL;const int maxx = 110;const LL INF = (1LL << 32);int n,q;LL d[maxx][maxx];void floyd() {for (int k = 0; k < n; k++) {for (int i = 0; i < n; i++) {if (d[i][k] == INF)continue;for (int j = 0; j < n; j++) {if (d[k][j] == INF)continue;d[i][j] = min(d[i][j], d[i][k] + d[k][j]);}}}}int main(){int u, v, c;scanf("%d %d", &n, &q);for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++)d[i][j] = i == j ? 0 : INF;}// 输入权值for (int i = 0; i < q; i++) {scanf("%d %d %d", &u, &v, &c);d[u][v] = c;}floyd();// 判断是否存在负环bool negative = false;for (int i = 0; i < n; i++)if (d[i][i] < 0)negative = true;if (negative)printf("NEGATIVE CYCLE\n");else {for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (d[i][j] == INF)printf("INF");elseprintf("%d", d[i][j]);printf("%c", j == n - 1 ? '\n' : ' ');}}}return 0;}

15.2 拓扑排序

题解

代码实现

#include <iostream>#include <cstdio>#include <vector>#include <queue>#include <list>#include <algorithm>using namespace std;const int maxx = 100010;const int INF = 1 << 30;vector<int>G[maxx];list<int>out;bool flag[maxx];int n;int indeg[maxx];void bfs(int s) {queue<int> q;q.push(s);while (!q.empty()) {int u = q.front(); q.pop();out.push_back(u);for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];indeg[v]--;if (indeg[v] == 0 && flag[v] == 0) {flag[v] = true;q.push(v);}}}}void tsort() {for (int i = 0; i < n; i++) {indeg[i] = 0;flag[i] = false;}for (int u = 0; u < n; u++) {for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];indeg[v]++;}}for (int u = 0; u < n; u++) {if (indeg[u] == 0 && flag[u] == 0)bfs(u);}for (list<int>::iterator it = out.begin(); it != out.end(); it++)printf("%d\n", *it);}int main(){int s, t, m;scanf("%d %d", &n, &m);for (int i = 0; i < m; i++) {scanf("%d %d", &s, &t);G[s].push_back(t);}tsort();return 0;}

15.4 树的直径

题解

代码实现

#include <iostream>#include <cstdio>#include <vector>#include <queue>using namespace std;const int maxx = 100010;const int INF = 1 << 30;class Edge {public:int t, w;Edge() {};Edge(int t ,int w):t(t), w(w) {};};vector<Edge> G[maxx];int n, d[maxx];bool vis[maxx];int cnt;void bfs(int s) {for (int i = 0; i < n; i++)d[i] = INF;queue<int> Q;Q.push(s);d[s] = 0;while (!Q.empty()) {int u = Q.front(); Q.pop();int len = G[u].size();for (int i = 0; i < len; i++) {Edge e = G[u][i];if (d[e.t] == INF) {d[e.t] = d[u] + e.w;Q.push(e.t);}}}}void solve() {// 从任选的结点s出发，选择距离s最远的结点tgtbfs(0);int maxv = 0, tgt = 0;for (int i = 0; i < n; i++) {if (d[i] == INF)continue;if (maxv < d[i]) {maxv = d[i];tgt = i;}}//从tgt出发，求结点tgt到最远结点的距离maxvbfs(tgt);maxv = 0;for (int i = 0; i < n; i++) {if (d[i] == INF)continue;maxv = max(maxv, d[i]);}printf("%d\n", maxv);}int main() {int s, t, w;scanf("%d", &n);for (int i = 0; i < n - 1; i++) {scanf("%d %d %d", &s, &t, &w);// 无向图G[s].push_back(Edge(t, w));G[t].push_back(Edge(s, w));}solve();return 0;}

15.5 最小生成树

题解

代码实现

第十七章动态规划略

第十八章数论

算法之经典数学知识

第十九章启发式搜索略

19.1 八皇后问题

19.2 九宫格拼图

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【操作指导 | 代码实现】挑战程序设计竞赛2：算法和数据结构

书籍封面

第一章 前言

1. 本人衷心建议

2. 本书涉及的所有题目以及对应的编号

3. AOJ 使用方法《挑战程序设计竞赛》

网址

注册

查找题目

做题

用 virtual judge 做 AOJ 的题目

第二章 算法与复杂度

2.4 算法的效率

2.4.1 复杂度的评估

2.4.2 大 O 表示法

2.4.3 复杂度的比较

第三章 初等排序

3.1 挑战问题之前——排序

3.2 插入排序法

题解

代码实现

3.3 冒泡排序法

题解

代码实现

3.4 选择排序法

题解

代码实现

3.6 希尔排序法

题解

代码实现

第四章 数据结构

4.1 挑战问题之前一一什么是数据结构

4.2 栈

题解

代码实现

4.3 队列

题解

代码实现

4.4 链 表

题解

代码实现

4.5 标准库的数据结构

4.5.2 stack

4.5.3 queue

4.5.4 vector

4.5.5 list

第五章 搜索

5.1 挑战问题之前一搜索

5.2 线性搜索

题解

代码实现

5.3 二分搜索

题解

代码实现

5.4 散列法

题解

代码实现

5.5 借助标准库搜索

5.5.1 迭 代 器

5.5.2 lower bound

5.6 搜索的应用——计算最优解

题解

代码实现

第六章 递归和分治法

6.1 挑战问题之前一递归与分治

6.2 穷 举 搜 索

题解

代码实现

6.3 科 赫 曲 线

题解

代码实现

第八章 树

8.2 有根树的表达

题解

代码实现

8.3 二叉树的表达

题解

代码实现

8.4 树的遍历

题解

第一章前言

第二章算法与复杂度

第三章初等排序

第四章数据结构

4.4 链表

第五章搜索

5.5.1 迭代器

第六章递归和分治法

6.2 穷举搜索

6.3 科赫曲线

第八章树

第九章二叉搜索树

第十章堆

第十一章动态规划

第十二章图

12.5 连通分量

第十三章加权图

第十四章高等数据结构

第十五章高等图算法

15.2 拓扑排序