Max Sum Plus Plus-HDU1024(dp)

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 ... 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

 

 

 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>

using namespace std;
#define N 1000010
#define INF 0xfffffff
int dp[N];
int M[N];
int Max;
int a[N];

int main()
{
    int m,n;
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            dp[i]=0;
            M[i]=0;
        }
        dp[0]=0;
        M[0]=0;
        for(int i=1;i<=m;i++)
        {
            Max=-INF;
            for(int j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1]+a[j],M[j-1]+a[j]);
                M[j-1]=Max;
                Max=max(Max,dp[j]);
            }
        }
        printf("%d\n",Max);
    }
    return 0;
}