Codeforces 574B Bear and Three Musketeers

1.英文题意

B. Bear and Three Musketeers

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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

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.

2.中文题意

　　现有一无向无重边图，求其中一个三元环，要求与该环相邻的边数总和最少，若无三元环输出“-1”，否则输出“与该环相邻的边数总和”。

3.思路

　　CF div2 B题Ｏ(n^3)可过，我的是O(n^2)，预处理求得每个顶点度数和，邻接矩阵和边集，每次检验边集中的一条边，是否有第三个点可与它构成环，更新“环上三个点度数和最小值-6”（不含3元环上度）。

4.实现

/*********************************
日期：2015-08-29
作者：matrix68
题号: Codeforces 574B Bear and Three Musketeers
**********************************/
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#define MAXN 4000
#define inf 99999
using namespace std;
typedef struct
{
int x;
int y;
} Node;

Node net[MAXN+10];//边集
int Du[MAXN+10];//度数和
bool matrix[MAXN+10][MAXN+10];//邻接矩阵

int main()
{
//    freopen("in.txt","r",stdin);
//     freopen("out.txt","w",stdout);
int n,m;
while(~scanf("%d%d",&n,&m))
{
memset(matrix,0,sizeof matrix);
memset(Du,0,sizeof Du);
for(int i=0; i<m; i++)
{
int a,b;
scanf("%d %d",&a,&b);
net[i].x=a;
net[i].y=b;
Du[a]++;
Du[b]++;
matrix[a][b]=matrix[b][a]=1;
}
int ans=inf;
bool flag=false;
//找环
for(int i=0; i<m; i++)
{
int a0=net[i].x;
int a1=net[i].y;
for(int j=1; j<=n; j++)
{
if(matrix[a1][j])
{
int a2=j;
if(matrix[a0][a2])
{
flag=true;
int tmp=Du[a0]+Du[a1]+Du[a2]-6;
ans=min(ans,tmp);
}
}
}
}
if(flag)
cout<<ans<<endl;
else
cout<<-1<<endl;
}
return 0;
}