POJ 2531 Network Saboteur(DFS)

Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10216		Accepted: 4885

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

3
0 50 30
50 0 40
30 40 0

Sample Output

90

题目大意： 有两个集合，求不同集合间的最大路径和。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<cmath>
#include<queue>
#define mod 9901
#define LL long long
using namespace std;
int n;
int a[32][32];
bool vis[32];
int ans;
void dfs(int s,int sum)
{
     vis[s]=true;//利用标记数组划分出两个集合
    for(int i=1;i<=n;i++)
    {
        if(vis[i])//假设当前的数在集合中减去当前数到该集合的距离
            sum-=a[s][i];
        else
            sum+=a[s][i];//否则加上
    }
    ans=max(sum,ans);
    for(int i=s+1;i<=n;i++)//枚举下一个点再增加集合
    {
        vis[i]=true;
        dfs(i,sum);
        vis[i]=false;
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        memset(vis,false,sizeof(vis));
        dfs(1,0);
    printf("%d
",ans);
    }
    return 0;
}