Codeforces Round #318 (Div. 2) B Bear and Three Musketeers （暴力）

算一下复杂度。发现可以直接暴。对于u枚举a和b，判断一下是否连边，更新答案。

#include<bits/stdc++.h>
using namespace std;

int n,m;
const int maxn = ;
#define PB push_back
vector<int> G[maxn];
bool g[maxn][maxn];
int deg[maxn];
const int INF = 0x3f3f3f3f;
int main()
{
    //freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(int i = ; i < m; i++){
        int a,b;scanf("%d%d",&a,&b);
        deg[a]++; deg[b]++;
        g[a][b] = g[b][a] = true;
        G[a].PB(b); G[b].PB(a);
    }
    int ans = INF;
    for(int u = ; u <= n; u++){
        int sz = G[u].size();
        for(int i = ; i < sz; i++){
            int a = G[u][i];
            for(int j = i+; j < sz; j++){
                int b = G[u][j];
                if(g[a][b]){
                    ans = min(ans,deg[u]+deg[a]+deg[b]-);
                }
            }
        }
    }
    if(ans<INF){
        printf("%d\n",ans);
    }else printf("-1\n");
    return ;
}