Max Sum Plus Plus

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(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. ^_^ 

InputEach 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. 
OutputOutput 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.

 题目意思为给一个长度为n的序列，给出一个整数m，在序列中取m段不相交的子序列，要求m子段的和最大值

 解题思路

设置dp[i][j]为前j个数分成i段的最大值，且最后由第j个数结尾，再设一个数组sum，sum[i][j]表示前j个数分成i段的最大值，不一定是以第j个数结尾，所以我们可以推出状态转移方程dp[i][j] = max(dp[i][j-1]+a[j], sum[i-1][j-1]+a[j]); 由于n比较大，不能开

那么大的二维数组所以需要优化，可以采用滚动数组，之后状态转移方程变为dp[j] = max(dp[j]+a[j], sum[j-1]+a[j]);

代码如下：

#include<stdio.h>
#include<iostream>
#include<string.h>
#define MIN  -1000000000
using namespace std;
const int maxn = 1e6 + 10;
int a[maxn];
int dp[maxn], sum[maxn];
int main()
{
    int m, n;
    while(scanf("%d%d", &m, &n) != EOF)
    {
        for(int i = 1; i <= n; i ++)
        {
            scanf("%d", &a[i]);
        }
        memset(dp, 0, sizeof(dp));
        memset(sum, 0, sizeof(sum));
        int maxx;
        for(int i = 1; i <= m; i ++)
        {
            maxx = MIN;
            for(int j = i; j <= n; j ++)
            {
                dp[j] = max(dp[j-1]+a[j], sum[j-1]+a[j]);
                sum[j-1] = maxx;
                maxx = max(maxx, dp[j]);
            }
        }
        //sum[n] = maxx;
        printf("%d\n", maxx);
        //cout<<endl;
    }
    return 0;
}