CF#318 (Div. 2)B. Bear and Three Musketeers 暴力 复杂度分析

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const int big=50000;
int max(int a,int b) {return a>b?a:b;};
int min(int a,int b) {return a<b?a:b;};
int flag[4005][4005],d[4005];
int main()
{
    int n,m,u,v;
    while(~scanf("%d %d",&n,&m))
    {
        memset(flag,0,sizeof(flag));
        memset(d,0,sizeof(d));
        for(int i=1;i<=m;i++)
            {
                scanf("%d %d",&u,&v);
                flag[u][v]=flag[v][u]=1;
                d[u]++;d[v]++;
            }
        int ans=inf;
        for(int i=1;i<=n;i++)
          for(int j=i+1;j<=n;j++)
             if(flag[i][j])
              for(int k=j+1;k<=n;k++)
                if(flag[i][k]&&flag[k][j])
              ans=min(ans,d[i]+d[j]+d[k]-6);
        printf("%d\n",ans==inf?-1:ans);
    }
    return 0;
}