06-4. How Long Does It Take (25)
时间限制 200 ms内存限制 65536 kB
代码长度限制 8000 B
判题程序 Standard 作者 CHEN, Yue
Given the relations of all the activities of a project, you are supposed to find the earliest completion time of the project.
Input Specification:
Each input file contains one test case. Each case starts with a line containing two positive integers N (<=100), the number of activity check points (hence it is assumed that the check points are numbered from 0 to N-1), and M, the number of activities. Then M lines follow, each gives the description of an activity. For the i-th activity, three non-negative numbers are given: S[i], E[i], and L[i], where S[i] is the index of the starting check point, E[i] of the ending check point, and L[i] the lasting time of the activity. The numbers in a line are separated by a space.
Output Specification:
For each test case, if the scheduling is possible, print in a line its earliest completion time; or simply output "Impossible".
Sample Input 1:
9 120 1 60 2 40 3 51 4 12 4 13 5 25 4 04 6 94 7 75 7 46 8 27 8 4
Sample Output 1:
18
Sample Input 2:
4 50 1 10 2 22 1 31 3 43 2 5
Sample Output 2:
Impossible
#include <iostream>#include <algorithm>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include<queue>#include<stack>#include<map>#include<set>using namespace std;#define lson rt<<1,l,MID#define rson rt<<1|1,MID+1,r//#define lson root<<1//#define rson root<<1|1#define MID ((l+r)>>1)typedef long long ll;typedef pair<int,int> P;const int maxn=3005;const int base=1000;const int inf=9999999999;const double eps=1e-5;int n,m;struct node{int to,cost;};vector<node> G[maxn];//邻接表int d[maxn];int earlist[maxn];//活动最早结束的数组bool tupo_sort()//拓扑排序 求最长的路径{stack<int> s;for(int i=0;i<n;i++){if(d[i]==0)s.push(i);}int cnt=0;//用于判断是否有环while(!s.empty()){int m=s.top();cnt++;s.pop();for(int i=0;i<G[m].size();i++){node tmp=G[m][i];if(--d[tmp.to]==0)s.push(tmp.to);if(earlist[m]+tmp.cost>earlist[tmp.to])//计算最长的路径{earlist[tmp.to]=earlist[m]+tmp.cost;}}}if(cnt<n)return false;//判断是否有环return true;}int main(){int i,j,k,t;cin>>n>>m;for(i=0;i<m;i++){node e;int s;cin>>s>>e.to>>e.cost;G[s].push_back(e);d[e.to]++;}if(tupo_sort())printf("%d\n",*max_element(earlist,earlist+n));//输出earlist数组的最大元素 elseputs("Impossible");return 0;}
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