Max Sum Plus Plus

这道题目是求m个不相交子串，使得这些子串的和最大，对于我来说这道题目挺难的，大体说一下思路吧，假设在前面已经有了i-1个子串（i<=m），那么循环从i开始（假设每个子串都最小只有一个数字，又因为所有的子串都是不相交的，第i个子串只能从第i个数字开始），开始进行判断，是这个数字单独构成一个子串最优，还是这个数字与它前面的数字构成一个子串最优（不太会写递推公式，只能这么表述了）

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S1, S2, S3, S4... Sx, ... Sn(1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx≤ 32767). We define a function sum(i, j) = Si+ ... + Sj(1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix≤ iy≤ jxor ix≤ jy≤ jxis not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^

Input Each test case will begin with two integers m and n, followed by n integers S1, S2, S3... Sn.

Process to the end of file.

Output Output the maximal summation described above in one line.

Sample Input

1 3 1 2 32 6 -1 4 -2 3 -2 3

Sample Output

68

Hint

Huge input, scanf and dynamic programming is recommended.

#include<stdio.h>#include<string.h>#include<algorithm>#define inf 0x3fffffff#define maxn 1000005using namespace std;int dp[maxn],pre[maxn],an[maxn];int main(){int m,n;while(scanf("%d%d",&m,&n)!=EOF){int ans;for(int i=1;i<=n;i++){scanf("%d",&an[i]);dp[i]=0;pre[i]=0;}dp[0]=0;pre[0]=0;for(int i=1;i<=m;i++){ans=-inf;for(int j=i;j<=n;j++){dp[j]=max(pre[j-1]+an[j],dp[j-1]+an[j]);pre[j-1]=ans;ans=max(dp[j],ans);}}printf("%d\n",ans);}return 0;}

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