HDU1024 最大m子段和

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27582    Accepted Submission(s): 9617

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 ... S_x, ... S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767). We define a function sum(i, j) = S_i + ... + S_j (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(i_m, j_m) maximal (i_x ≤ i_y ≤ j_x or i_x ≤ j_y ≤ j_x 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(i_x, j_x)(1 ≤ x ≤ m) instead. ^_^

 

Input

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

 

 题意：

求最大m子段和。

代码：

//最大m子段和递推公式:dp[i][j]=max(dp[i][j-1]+num[j],dp[i-1][t]+num[j]) (i<=t<=j-1)
//优化：因为每次都要求一个最大的dp[i-1][t](i<=t<=j-1),可以每次求出来dp[i][j]时保存
//这个值，到下一次时直接用就行了，这样每次就只用到了dp[i][j]和dp[i][j-1],可以省去一维。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1000006;
const int inf=0x7fffffff;
int dp[maxn],fdp[maxn],num[maxn];
int main()
{
    int m,n,maxnum;
    while(scanf("%d%d",&m,&n)==2){
        for(int i=1;i<=n;i++) scanf("%d",&num[i]);
        memset(dp,0,sizeof(dp));
        memset(fdp,0,sizeof(fdp));
        for(int i=1;i<=m;i++){
            maxnum=-inf;
            for(int j=i;j<=n;j++){
                dp[j]=max(dp[j-1]+num[j],fdp[j-1]+num[j]);
                fdp[j-1]=maxnum;
                maxnum=max(maxnum,dp[j]);
            }
        }
        printf("%d
",maxnum);
    }
    return 0;
}