HDU1024 Max Sum Plus Plus ——dp

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30682    Accepted Submission(s): 10828

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
Hint
Huge input, scanf and dynamic programming is recommended.
 

 

Author

JGShining（极光炫影）

题意：给一个数字序列，求其中的m段的和，使这个和最大。这m段互不相交。

分析：使用动态规划。dp[i][j]表示前j个数字中的i段最大和，并且它的最后一段是以a[j]结尾的。状态转移方程为：

dp[i][j] = max(dp[i][j-1]+a[j], dp[i-1][k]+a[j])，其中k的取值范围是：i-1<= k <=j-1

解释：1、dp[i][j-1]是前j-1个元素分成了i段，这个时候只需要将a[j]加在最后一段即可

2、dp[i-1][k]是前k个元素分成了i-1段，这个时候将a[j]这一个数作为第i段。

代码：

#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;#define N 1000005typedef long long ll;ll dp[2][N];ll a[N];int main(){int n, m;while(~scanf("%d", &m)){scanf("%d", &n);for(int i = 1; i <= n; i++)scanf("%lld", &a[i]);for(int i = 0; i <= n; i++)dp[0][i] = dp[1][i] = 0;int index = 0;for(int i = 1; i <= m; i++){dp[index][i] = dp[1-index][i-1] + a[i];ll M = dp[1-index][i-1];for(int j = i+1; j <= n-m+i; j++){M = max(M, dp[1-index][j-1]);dp[index][j] = max(dp[index][j-1], M) + a[j];}index = 1 - index;}index = 1 - index;ll ans = dp[index][m];for(int i = m+1; i <= n; i++)ans = max(ans, dp[index][i]);printf("%lld\n", ans);}return 0;} 

注意代码中的优化，如果k值每次都从i-1到j-1遍历一遍的话，会超时。

因此用一个M表示dp[i-1][i-1]到dp[i-1][j-1]的最大值，j每次+1时都更新M为从i-1到j-1的最大值。