Max Sum Plus Plus

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. ^_^

 

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 the maximal summation described above in one line.

 

 

 

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 }