HDU 1024 Max Sum Plus Plus （最大和子序列增强版：求规定下标内的最大值）

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17906    Accepted Submission(s): 5864

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<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define inf 999999999
//using namespace std;
const int MAX = 1000005;
int max(int a,int b)
{
    return a>b?a:b;
}
int a[MAX];
int left[MAX];
int right[MAX];
int main()
{
    int n,m,i,j;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(left,0,sizeof(left));
        memset(right,0,sizeof(right));
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        long long count;
        for(i=1;i<=m;i++)
        {
            count=-inf;
            for(j=i;j<=n;j++)
            {
                left[j]=max(left[j-1],right[j-1])+a[j];
                right[j-1]=count;
                if(left[j]>count)
                    count=left[j];
            }
        }
        printf("%I64d\n",count);
    }
    return 0;
}