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 = 4001;
#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 = 0; 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 = 1; u <= n; u++){int sz = G[u].size();for(int i = 0; i < sz; i++){int a = G[u][i];for(int j = i+1; j < sz; j++){int b = G[u][j];if(g[a][b]){ans = min(ans,deg[u]+deg[a]+deg[b]-6);}}}}if(ans<INF){printf("%d\n",ans);}else printf("-1\n");return 0;
}

转载于:https://www.cnblogs.com/jerryRey/p/4770261.html

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