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

     return ;
 }