Network Saboteur-POJ2531

Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14300		Accepted: 7023

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

Source

Northeastern Europe 2002, Far-Eastern Subregion

题目意思：首先的搞清楚矩阵表示的意思

0 50 30   //表示1-1的距离为0，1-2的距离为50，1-3的距离为30

50 0 40//表示2-1的距离为50，对称。。。。。

30 40 0

将这些点任意分成两部分，求其中一部分点到另外一部分点的权值的和的最大值

代码如下:

#include<iostream>
#include<string.h>
#include<cstdio>
#include<algorithm>
using namespace std;
int maxn,matrix[25][25],vis[25],n;
void dfs(int index,int sum)
{
	vis[index]=1;
	for(int i=0;i<n;i++){
		if(vis[i]){//如果为同一个集合，则之间的距离会为零,所以传过来的sum值需要减去这一部分 
			sum-=matrix[index][i];
		}
		else{
			sum+=matrix[index][i];
		}
	}
	maxn=max(maxn,sum);
	for(int i=index+1;i<n;i++){
		dfs(i,sum);
		vis[i]=0;	
	}
}
int main()
{
	while(~scanf("%d",&n)){
		memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				scanf("%d",&matrix[i][j]);
			} 
		}
		maxn=0;
		dfs(0,0);
		printf("%d\n",maxn);
	}
	return 0;
}