HDU 1024：Max Sum Plus Plus（DP）

http://acm.hdu.edu.cn/showproblem.php?pid=1024

Max Sum Plus Plus

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_xor 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.

 

题意：将一串数字分成m段子序列，求最大可能达到的总和。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 #define N 1000005
 #define INF -1000000000
 int num[N],dp[N],mmax[N];
 /*
 状态DP[i][j]
 前j个数可以分成i组的和的最大值
 DP[i][j]=max(dp[i][j-1]+num[j], max( dp[i-1][k] )+num[j]),0<k<j;
 dp[i][j-1]是把num[j]加入到前面的一个组，
 max(dp[i-1][k])+num[j]是把num[j]重新分在另一个组里面，
 而max(dp[i-1][k])是分成i-1个组时候的和的最大值
 开一个mmax数组每次都可以保存分成i-1个组的时候使用前j-1个数时候最大值
 时间复杂度O(mn)
 */
 int main()
 {
     int n,m;
     while(~scanf("%d%d",&m,&n)){
         for(int i=;i<=n;i++){
             scanf("%d",num+i);
         }
         memset(dp,,sizeof(dp));
         memset(mmax,,sizeof(mmax));
         int ans;
         for(int i=;i<=m;i++){
             ans=INF;
             for(int j=i;j<=n;j++){
                 dp[j]=max(dp[j-],mmax[j-])+num[j];
                 mmax[j-]=ans;
                 ans=max(dp[j],ans);
             }
         }
         cout<<ans<<endl;
     }
     return ;
 }

2016-06-21