HDU 1024：Max Sum Plus Plus（DP）

Posted 2020-07-27 Shadowdsp

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

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24675    Accepted Submission(s): 8478

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.

 

题意：将一串数字分成m段子序列，求最大可能达到的总和。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<iostream>
 5 using namespace std;
 6 #define N 1000005
 7 #define INF -1000000000
 8 int num[N],dp[N],mmax[N];
 9 /*
10 状态DP[i][j]
11 前j个数可以分成i组的和的最大值
12 DP[i][j]=max(dp[i][j-1]+num[j], max( dp[i-1][k] )+num[j]),0<k<j;
13 dp[i][j-1]是把num[j]加入到前面的一个组，
14 max(dp[i-1][k])+num[j]是把num[j]重新分在另一个组里面，
15 而max(dp[i-1][k])是分成i-1个组时候的和的最大值
16 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
17 时间复杂度O(mn)
18 */
19 int main()
20 {
21     int n,m;
22     while(~scanf("%d%d",&m,&n)){
23         for(int i=1;i<=n;i++){
24             scanf("%d",num+i);
25         }
26         memset(dp,0,sizeof(dp));
27         memset(mmax,0,sizeof(mmax));
28         int ans;
29         for(int i=1;i<=m;i++){
30             ans=INF;
31             for(int j=i;j<=n;j++){
32                 dp[j]=max(dp[j-1],mmax[j-1])+num[j];
33                 mmax[j-1]=ans;
34                 ans=max(dp[j],ans);
35             }
36         }
37         cout<<ans<<endl;
38     }
39     return 0;
40 }

 

2016-06-21