hdu 1024 Max Sum Plus Plus

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 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^
 


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


题意： 求m个最大子序列的和的最大值。
状态转移方程：dp[i][j]表示以i为结尾元素的j个子段的数和
dp[i][j]=max(dp[i-1][j]+a[i],dp[i-k][j-1]+a[i]);其中(j-1<=k<=n-m+1)


max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。
我们可以在每次计算dp[i][j]的时候记录下前j个的最大值 

用数组保存下来  下次计算的时候可以用，这样时间复杂度为 n^2.

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<queue>
#define INF 1<<30;;
#define N 1000005
using namespace std;


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