最大m字段和

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30233    Accepted Submission(s): 10634

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
dp[i][j]表示前j项中i个字段和的最大值
dp[i][j]=max(dp[i][j-1]+dp[i-1][t])+num[j]
#include<iostream>//最大m字段和问题
#include<stdio.h>
#include<string.h>
using namespace std;
int val[1000001],ans[1000001],lans[1000001];
int max(int a,int b)
{
    return a>b?a:b;
    }
int main()
{
    int n,m,i,j;
    //cout<<max(7,8);
    while(cin>>n>>m)
    {
        memset(ans,0,sizeof(ans));
        memset(lans,0,sizeof(lans));
        for(i=1;i<=m;i++)
        {
            scanf("%d",&val[i]);
        }
        long   maxx;
        for(i=1;i<=n;i++)
        {
            maxx=-99*1000000;
            for(j=i;j<=m;j++)
            {
                ans[j]=max(ans[j-1],lans[j-1])+val[j];//当前的最大值是前一项的最大值和上一个状态较大的值加上目前的值
                lans[j-1]=maxx ;//修改前一项的值，不能和前一行互换
                if(ans[j]>maxx )
                maxx=ans[j];
                }
            }
        cout<<maxx<<endl;
        }
    return 0;
}