B. Bear and Three Musketeers

最新推荐文章于 2020-12-13 20:42:20 发布

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原文链接：http://www.cnblogs.com/NaCl/p/4777070.html

Richelimakieu希望从一群战士中选择三位相互认识的战士来帮助他对抗坏人，但必须确保他们之间的认识度之和最小。通过分析战士间的认识关系，找出满足条件的最佳组合。

Do you know a story about the three musketeers? Anyway, you will learn about its origins now.

Richelimakieu is a cardinal in the city of Bearis. He is tired of dealing with crime by himself. He needs three brave warriors to help him to fight against bad guys.

There are n warriors. Richelimakieu wants to choose three of them to become musketeers but it's not that easy. The most important condition is that musketeers must know each other to cooperate efficiently. And they shouldn't be too well known because they could be betrayed by old friends. For each musketeer his recognition is the number of warriors he knows, excluding other two musketeers.

Help Richelimakieu! Find if it is possible to choose three musketeers knowing each other, and what is minimum possible sum of their recognitions.

Input

The first line contains two space-separated integers, n and m (3 ≤ n ≤ 4000, 0 ≤ m ≤ 4000) — respectively number of warriors and number of pairs of warriors knowing each other.

i-th of the following m lines contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). Warriors ai and bi know each other. Each pair of warriors will be listed at most once.

Output

If Richelimakieu can choose three musketeers, print the minimum possible sum of their recognitions. Otherwise, print "-1" (without the quotes).

Sample test(s)

input

output

input

output

-1
这题其实就是问你能不能找到三个彼此认识的人，然后那三个人认识的人数和要最少。。。暴力枚举三个成环的情况，然后更新最少的认识人数和

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
vector<int>V[5000]; 
int f[5000][5000];
int main()
{
    int n,m;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        int x,y;
        for(int i=1;i<=n;i++)
        V[i].clear();
        memset(f,0,sizeof(f));
        for(int i=0;i<m;i++)
        {
            scanf("%d %d",&x,&y);
            V[x].push_back(y);
            V[y].push_back(x);
            f[x][y]=f[y][x]=1;
        }
        int ans=100000000;
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<V[i].size();j++)
            {
                int k=V[i][j];
                for(int l=0;l<V[k].size();l++)
                {
                    int a=V[k][l];
                    if(f[i][k]&&f[k][a]&&f[a][i])
                    {
                        int sum=V[i].size()+V[k].size()+V[a].size()-6;
                      if(sum<ans)
                      {
                          ans=sum;
                          flag=1;
                      }
                    }
                }
            }
        } 
        if(flag)
    printf("%d\n",ans);
    else
    printf("-1\n");
    }
    return 0;
}