Fellow me on GitHub

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 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^
 

 

Input

Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
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
 1 //2017-04-04
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 const int N = 1000005;
10 const int inf = 0x3f3f3f3f;
11 int s[N], dp[N], mx[N];
12 
13 int main()
14 {
15     int n, m, maxx;
16     while(scanf("%d%d", &m, &n)!=EOF)
17     {
18         for(int i = 1; i <= n; i++)
19         {
20             scanf("%d", &s[i]);
21             dp[i] = 0;
22             mx[i] = 0;
23         }
24         dp[0] = mx[0] = 0;
25         for(int i = 1; i <= m; i++)
26         {
27             maxx = -inf;
28             for(int j = i; j <= n; j++)
29             {
30                 dp[j] = max(dp[j-1]+s[j], mx[j-1]+s[j]);
31                 mx[j-1] = maxx;
32                 maxx = max(maxx, dp[j]);
33             }
34         }
35         cout<<maxx<<endl;
36     }
37 
38     return 0;
39 }

 

posted @ 2017-04-04 10:05  Penn000  阅读(276)  评论(0编辑  收藏  举报