Max Sum Plus Plus m段连续子序列最大和

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

算是最大连续子序列和的加强版吧

状态：dp[i][j]表示前j个元素组成i组子序列和的最大值

状态转移：dp[i][j] = max(dp[i][j-1], max(dp[i-1][k])) + a[j] ，注意这里 i-1 <= k >= j-1

但是 j <= n <= 10000000，而且m的范围不知道，所以很有可能超内存...

看了网上的做法，说用什么滚动数组？记录前一组的最大值，所以是不断更新的

#include <iostream>  
#include <algorithm>  
#include <cstdio>  
#include <cstring>

using namespace std;  
const int N = 1e6 + 3;

int dp[N], a[N], pre[N];  //pre数组记录前一段和的最大值

int main()  
{      
    int m, n;
    while(~scanf("%d%d", &m, &n))
    {
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        memset(pre, 0, sizeof(pre));
        dp[0] = 0;
        int p;
        for(int i = 1; i <= m; i++)
        {
            p = -1e9;
            for(int j = i; j <= n; j++)  //这里j一定要从i开始，因为每一组的元素数量肯定至少是1...
            {
                dp[j] = max(dp[j-1], pre[j-1]) + a[j];
                pre[j-1] = p;  //记录下来，为i+1组使用，使用后更新，为下一组使用
                p = max(p, dp[j]);
            }
        }
        printf("%d\n", p);
    }
    return 0;  
}