Max Sum Plus Plus_DP

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25572    Accepted Submission(s): 8850

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

Hint

Huge input, scanf and dynamic programming is recommended.

由于N比较大，用二维数组会爆掉

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int inf=0x7777777;
const int N=1000005;
int n,m;
int maxx[N],a[N],dp[N];
int main()
{
    int mm;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&a[i]);
            maxx[i]=0,dp[i]=0;
        }
        maxx[0]=0,dp[0]=0;
        for(int i=1;i<=n;i++)
        {
            mm=-inf;
            for(int j=i;j<=m;j++)
            {
                dp[j]=max(dp[j-1]+a[j],maxx[j-1]+a[j]);//dp在是否取a[j]的两种情况取最大值
                maxx[j-1]=mm;//maxx记录到当前位置最大值
                mm=max(mm,dp[j]);//更新最大值
            }
        }
        cout<<mm<<endl;
    }
    return 0;
}