Max Sum Plus Plus（n个元素m段连续子段和的最大值）

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 S 1, S 2, S 3, S 4 ... 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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(im, 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 S 1, S 2, S 3 ... 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

DP动态规划思路

题意：求n个元素数组的m段连续子段和的最大值

dp[i][j]=前j个数选择i段的最大值

状态转移：dp[i][j]=max(dp[i][j-1]  ,  max{dp[i-1][k]}  )+a[j]  (i-1<=k<=j-1)

第J个元素可以和前面的连接在一起，构成i段，或者独自成为一段，

如果是独自成为一段的话，那么前j-1个数必须组合出i-1段，选择多种情况里的最大值，

最后由于数据较大，所以要压缩空间，使用滚动数组

AC参考代码：

PS：大量不能用cin和cout

加油变强！！！！！！！

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<iostream>
using namespace std;
const int maxn=1000005;
int str[maxn],pre[maxn];
int dp[maxn];
int main()
{
    int n,m;
    while(~scanf("%d%d",&m,&n))
    {
        int ans;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",str+i);
            dp[i]=pre[i]=0;
        }
        dp[0]=pre[0]=0;
        for(int i=1;i<=m;i++)
        {
            ans=-999999999;
            for(int j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1],pre[j-1])+str[j];
                pre[j-1]=ans;
                ans=max(dp[j],ans);
            }
        }
       printf("%d\n",ans);
    }
    return 0;
}