POJ题目2531Network Saboteur（枚举）

Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9527		Accepted: 4529

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

题目大意：把n个点分成两部分，是两部分的相连的权值最大，求这个最大值

思路：总共有n个点，每个点都有两种情况，再减去全是第一组和全是第二组的俩种情况，总共2^(n-1)中情况，用a数组01表示每个点的状态，枚举所有情况就好，最多不会超过20个点

ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
int a[22],map[22][22];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int i,j,ans=0,m,k;
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
				scanf("%d",&map[i][j]);
		m=1<<(n-1);
		memset(a,0,sizeof(a));
		for(i=0;i<m;i++)
		{
			a[0]++;
			int sum=0;
			for(j=0;j<n;j++)
			{
				if(a[j]==2)
				{
					a[j]=0;
					a[j+1]++;
				}
				else
					break;
			}
			for(j=0;j<n;j++)
			{
				if(!a[j])
					continue;
				for(k=0;k<n;k++)
				{
					if(!a[k])
						sum+=map[j][k];
				}
			}
			ans=max(sum,ans);
		}
		printf("%d\n",ans);
	}
}