题干：
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).
Examples
Input
5 6 1 2 1 3 2 3 2 4 3 4 4 5
Output
2
Input
7 4 2 1 3 6 5 1 1 7
Output
-1
Note
In the first sample Richelimakieu should choose a triple 1, 2, 3. The first musketeer doesn't know anyone except other two musketeers so his recognition is 0. The second musketeer has recognition 1 because he knows warrior number 4. The third musketeer also has recognition 1 because he knows warrior 4. Sum of recognitions is 0 + 1 + 1 = 2.
The other possible triple is 2, 3, 4 but it has greater sum of recognitions, equal to 1 + 1 + 1 = 3.
In the second sample there is no triple of warriors knowing each other.
解题报告：
显然，2000ms的时限，4000的数据量，考虑是否o(n^2)可以解决。（其实这题直接枚举三个点，o(n^3)也可以过？）怎么优化到平方级别呢？其实也很简单，就是第一层循环遍历m条边就好了。因为你三层for枚举顶点的时候，其实前两层有很多是浪费掉了的，因为有很多顶点的之间根本没有边，而你都遍历了一遍，所以我们都记录下来边了，为什么不直接遍历边呢？这样我们就有两个顶点了呀，然后再来一层循环遍历每一个点就好了。注意这种暴力的题一定别忘判断一下是否遍历到了自己，比如【CodeForces - 761C】Dasha and Password 这道题，最开始就忘了判断是否遍历到了同一个字符串。
AC代码：
#include<bits/stdc++.h>
using namespace std;
const int MAX = 4000 + 5 ;
int du[MAX];
bool maze[MAX][MAX];
int head[MAX];
struct Edge {
int fm,to;
int w;
}e[50000 + 5];
int top;
void add(int u,int v) {
e[++top].fm = u;
e[top].to = v;
}
int main()
{
int n,m,u,v;
cin>>n>>m;
for(int i = 1; i<=m; i++) {
scanf("%d%d",&u,&v);
add(u,v);add(v,u);
du[u]++;du[v]++;
maze[u][v] = maze[v][u] = 1;
}
int ans = 0x3f3f3f3f;
for(int i = 1; i<=top; i++) {
for(int j = 1; j<=n; j++) {
if(j == e[i].fm || j == e[i].to) continue;
if(maze[e[i].fm][j] && maze[e[i].to][j]) {
ans = min(ans,du[e[i].to] + du[e[i].fm] + du[j]);
}
}
}
if(ans == 0x3f3f3f3f) puts("-1");
else printf("%d\n",ans-6);
return 0 ;
} AC代码2：（其实也不是拓扑排序啦，就是用vector表示了二维数组，并且，由第二个点找第三个点的时候直接在第一个点所连的边中找就可以了。但是那个c函数其实也有点消耗时间？如果直接用个bool类型的maze二维数组来表示无向图两点之间是否相连就更快了）