Network Saboteur FZU - 1346

Network Saboteur

 FZU - 1346 

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
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

题解+解析:

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include <cstring>
#include <string>
#define INF 0x7f7f7f7f
typedef long long ll;
using namespace std;
bool vis[50];
int a[30][30];
int ans = -1;
int t;
void dfs(int c)
{
	if (c == t)//分配完节点,计算距离。
	{
		int count = 0;
		for (int i = 0; i < t; i++)
		{
			if (vis[i])
			{
				for (int j = 0; j < t; j++)
				{
					if (!vis[j])
					{
						count += a[i][j];
					}
				}
			}
		}
		if (count > ans)
		{
			ans = count;
			return;
		}
		else
		{
			return;
		}
	}
	
	vis[c] = true;//每个节点有两种可能
	dfs(c+1);
	if (c==0)//把第一个节点放进另一个集合里是重复的,剪枝
	{
		return;
	}
	vis[c] = false;
	dfs(c+1);
}
int main()
{
	while (scanf("%d", &t) != EOF)
	{
		ans = -1;
		memset(vis, 0, sizeof(vis));
		for (int i = 0; i < t; i++)
		{
			for (int j = 0; j < t; j++)
			{
				scanf("%d", &a[i][j]);
			}
		}
		dfs(0);//分配节点。
		printf("%d\n", ans);
	}
	
	return 0;
}


posted @ 2018-04-18 21:46  focus5679  阅读(97)  评论(0编辑  收藏  举报