Max Sum Plus Plus HDU - 1024（DP）

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. ^_
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.

题意：
n个数字，m个子段和的最大值。

思路：
最大子段和的升级版。
最大子段和的方程为dp[i] = max(dp[i - 1],0) + a[i]。dp[i]代表末尾数字为a[i]的最大子段和。
本题因为是多个子段，所以多加一维状态是dp[i][j]，有i段且结尾数字为a[j]的结果。
dp[i][j] = max(dp[i][j - 1],max{dp[i-1][k]}) + a[i]。 i -1 ≤ k ≤ j - 1。
省略一维数组，并且用一个数组pre记录最大值前面j-1个数的最大值。

#include 
#include 
#include 

using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 7;
int dp[maxn],pre[maxn];
int a[maxn];

int main()
{
    int m,n;
    while(~scanf("%d%d",&m,&n))
    {
        memset(dp,0,sizeof(dp));memset(pre,0,sizeof(pre));
        for(int i = 1;i <= n;i++)scanf("%d",&a[i]);
        int ans = -INF;
        for(int i = 1;i <= m;i++)
        {
            ans = -INF;
            for(int j = i;j <= n;j++)
            {
                dp[j] = max(dp[j - 1],pre[j - 1]) + a[j];
                pre[j - 1] = ans;
                ans = max(ans,dp[j]);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}