# POJ 2531 Network Saboteur (DFS)

Network Saboteur

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 12453 Accepted: 6009

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

### AC 代码

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<queue>
#include<string.h>
using namespace std;
int a[30][30];      //map
bool con[30];       //各点分组情况
int ans,n;

void dfs(int id,int tmp)    //将id移到右边，计算流量
{
int old=tmp;
con[id]=true;
for(int i=0; i<n; i++)  //同组减，异组加
{
if(con[i])tmp-=a[i][id];
else tmp+=a[i][id];
}
ans=max(ans,tmp);
for(int i=id+1; i<n; i++)   //搜索后面的点
if(!con[i]&&tmp>old)
dfs(i,tmp);
con[id]=false;
}

int main()
{
while(~scanf("%d",&n))
{
ans=0;
memset(con,false,sizeof(con));
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
scanf("%d",&a[i][j]);
for(int i=0; i<n; i++)
dfs(i,0);
printf("%d\n",ans);
}
return 0;
}