(搜索) poj 2531

时间：2015-02-12 13:57:24 阅读：151 评论：0 收藏：0 [点我收藏+]

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Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9531		Accepted: 4533

Description

A university network is composed of N computers. System administrators gathered information on the traffic between nodes, and carefully divided the network into two subnetworks in order to minimize traffic between parts.
A disgruntled computer science student Vasya, after being expelled from the university, decided to have his revenge. He hacked into the university network and decided to reassign computers to maximize the traffic between two subnetworks.
Unfortunately, he found that calculating such worst subdivision is one of those problems he, being a student, failed to solve. So he asks you, a more successful CS student, to help him.
The traffic data are given in the form of matrix C, where Cij is the amount of data sent between ith and jth nodes (Cij = Cji, Cii = 0). The goal is to divide the network nodes into the two disjointed subsets A and B so as to maximize the sum ∑Cij (i∈A,j∈B).

Input

The first line of input contains a number of nodes N (2 <= N <= 20). The following N lines, containing N space-separated integers each, represent the traffic matrix C (0 <= Cij <= 10000).
Output file must contain a single integer -- the maximum traffic between the subnetworks.

Output

Output must contain a single integer -- the maximum traffic between the subnetworks.

Sample Input

Sample Output

Source

Northeastern Europe 2002, Far-Eastern Subregion

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<queue>
using namespace std;
int n,a[30][30],ans,vis[30];
void dfs(int x)
{
    vis[x]=1;
    int cnt=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(vis[i]&&!vis[j])
                cnt+=a[i][j];
        }
    }
    if(cnt>ans)
        ans=cnt;
    for(int i=x+1;i<n;i++)
    {
        dfs(i);
        vis[i]=0;
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                scanf("%d",&a[i][j]);
        memset(vis,0,sizeof(vis));
        dfs(0);
        printf("%d\n",ans);
    }
    return 0;
}

(搜索) poj 2531

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原文地址：http://www.cnblogs.com/a972290869/p/4287798.html

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