杭电1024Max Sum Plus Plus动态规划

题目：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): 30977    Accepted Submission(s): 10914

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 32 6 -1 4 -2 3 -2 3

 

Sample Output

68
动态规划问题，n个数字，分为m个不相交的子段并使它们的和最大
代码如下：
#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>using namespace std;const int N = 1000010;int max(int a, int b){if (a > b)return a;elsereturn b;}int value[N];int need[N];int n, v, w, m;int S[N];int dp[N];int temp[N];int Sum;int main(){int i, j, k;while (scanf("%d %d", &m, &n) != EOF){for (i = 1; i <= n; i++){scanf("%d", &S[i]);dp[i] = 0;temp[i] = 0;}dp[0] = 0;temp[0] = 0;for (i = 1; i <= m; i++){Sum = -9999999;for (j = i; j <= n; j++){dp[j] = max(dp[j - 1] + S[j], temp[j - 1] + S[j]);temp[j - 1] = Sum;Sum = max(Sum, dp[j ]);}}cout << Sum << endl;}return 0;}