HDU1024（DP）

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27976    Accepted Submission(s): 9749

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.

 

 

Author

JGShining（极光炫影）

 

http://www.cnblogs.com/kuangbin/archive/2011/08/04/2127085.html

//2017-04-04
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1000005;
const int inf = 0x3f3f3f3f;
int s[N], dp[N], mx[N];

int main()
{
    int n, m, maxx;
    while(scanf("%d%d", &m, &n)!=EOF)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &s[i]);
            dp[i] = 0;
            mx[i] = 0;
        }
        dp[0] = mx[0] = 0;
        for(int i = 1; i <= m; i++)
        {
            maxx = -inf;
            for(int j = i; j <= n; j++)
            {
                dp[j] = max(dp[j-1]+s[j], mx[j-1]+s[j]);
                mx[j-1] = maxx;
                maxx = max(maxx, dp[j]);
            }
        }
        cout<<maxx<<endl;
    }

    return 0;
}