Hdoj 1024 Max Sum Plus Plus 【DP】

Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18861 Accepted Submission(s): 6205

Problem Description

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 ≤ jx or ix ≤ jy ≤ jx is 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 3
2 6 -1 4 -2 3 -2 3

Sample Output

6
8

Hint

Huge input, scanf and dynamic programming is recommended.

Author

JGShining（极光炫影）
题意：求在给出的n个数中找出m个连续的子序列，使得各子序列之和最大。
解题链接
代码(不是太懂，什么时候懂了再来补一个详细解释吧)：

#include <cstdio>
#include <iostream>
#include <cstring>
const int M = 1e6+5;
#define LL __int64
using namespace std;

LL pre[M], num[M];

inline LL max(LL A, LL B){
    return A > B? A:B;
}

void DP(int m, int n){
    LL temp; //temp就是DP[I][J]
    for(int i = 1; i <= m; ++ i){
        temp = 0;
        for(int j = 1; j <= i; ++ j) temp += num[j];
        pre[n] = temp;
        for(int j = i+1; j <= n; ++ j){
            temp = max(temp, pre[j-1])+num[j]; //pre[j-1]就是max（dp[i-1][t]）(i-1 <= t <= i-1);
            pre[j-1] = pre[n];
            pre[n] = max(temp, pre[n]);
        }
    }
}

int main(){
    int n, m;
    while(scanf("%d%d", &m, &n) == 2){
        for(int i = 1; i <= n; ++ i){
            scanf("%I64d", num+i); pre[i] = 0;
        }
        DP(m, n);
        printf("%I64d\n", pre[n]);
    }
    return 0;
}