Maximum sum(poj 2479 最大m子段和)

最新推荐文章于 2019-08-08 09:28:23 发布

原创最新推荐文章于 2019-08-08 09:28:23 发布 · 1.3k 阅读

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本博客探讨了一个关于连续子段数和最大值的问题，通过实例分析和代码实现来解决问题，涉及算法和编程技巧。

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1、http://poj.org/problem?id=2479

2、题目大意：

给定一串数字，求连续的两个不相交的子段的数和最大，由题目不难看出此题求得是两个子段的数和

3、思路：跟上一篇博客hdu 1024 一样的想法，只是稍微处理一下题目，就不难看出类型一样

4、题目：

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28823		Accepted: 8809

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

4、代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 50005
int a[N];
int now[N];
int pre[N];
int main()
{
    int t,n,maxx;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
        memset(now,0,sizeof(now));
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=2;i++)
        {
            maxx=-1*N*10000;
            for(int j=i;j<=n;j++)
            {
                now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);
                pre[j-1]=maxx;
                if(now[j]>maxx)
                maxx=now[j];
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}
/*
1

10
1 -1 2 2 3 -3 4 -4 5 -5
*/