Maximum sum（双向dp）

最新推荐文章于 2023-02-25 22:41:55 发布

原创最新推荐文章于 2023-02-25 22:41:55 发布 · 590 阅读

1 ·

CC 4.0 BY-SA版权

ACM_动态规划专栏收录该内容

14 篇文章

订阅专栏

本文介绍了一个关于计算最大子数组和的问题，通过动态规划的方法解决了一组整数中连续子数组的最大和问题，并提供了一个AC代码示例，该示例使用了两个动态规划数组来分别从左到右和从右到左进行计算。

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 38077		Accepted: 11902

Description

Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:

$d(A)=\max_{1\leq s_1\leq t_1<s_2\leq t_2\leq n}\left\{\sum_{i=s_1}^{t_1}a_i+\sum_{j=s_2}^{t_2}a_j\right\}$

Your task is to calculate d(A).

Input

The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.

Output

Print exactly one line for each test case. The line should contain the integer d(A).

Sample Input

1

10
1 -1 2 2 3 -3 4 -4 5 -5

Sample Output

AC代码：

#include <iostream>
#include<cmath>
using namespace std;
#define inf 0xfffffff
int dp[50010],dp1[50010],a[50010],sm[50010];
int main(){
	int t,n,i;
	int sum,temp;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(i=1;i<=n;i++)
		scanf("%d",&a[i]);
		temp=inf;
		dp[1]=a[1];
		dp1[n]=a[n];
		for(i=2;i<=n;i++)
		dp[i]=max(a[i],dp[i-1]+a[i]);		
		for(i=n-1;i>=1;i--)
		dp1[i]=max(a[i],dp1[i+1]+a[i]);
		sm[n]=dp1[n];
		for(i=n-1;i>=1;i--)
		sm[i]=max(sm[i+1],dp1[i]);
		sum=-inf;
		for(i=2;i<=n;i++)
		sum=max(sum,dp[i-1]+sm[i]);
		printf("%d\n",sum);
	}
	return 0;
}