Codeforces 489D 搜索

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, 本文介绍了一个算法问题，即在一个给定的有向图中寻找特定的四边形结构——Damn Rhombus。通过两次广度优先搜索（BFS）并记录节点访问次数的方法，实现了对这种特定结构的有效计数。

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传送门：http://codeforces.com/contest/489/problem/D

D. Unbearable Controversy of Being

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Tomash keeps wandering off and getting lost while he is walking along the streets of Berland. It's no surprise! In his home town, for any pair of intersections there is exactly one way to walk from one intersection to the other one. The capital of Berland is very different!

Tomash has noticed that even simple cases of ambiguity confuse him. So, when he sees a group of four distinct intersections a, b, c and d, such that there are two paths from a to c — one through b and the other one through d, he calls the group a "damn rhombus". Note that pairs (a, b), (b, c), (a, d), (d, c) should be directly connected by the roads. Schematically, a damn rhombus is shown on the figure below:

$\text{[math]}$

Other roads between any of the intersections don't make the rhombus any more appealing to Tomash, so the four intersections remain a "damn rhombus" for him.

Given that the capital of Berland has n intersections and m roads and all roads are unidirectional and are known in advance, find the number of "damn rhombi" in the city.

When rhombi are compared, the order of intersections b and d doesn't matter.

Input

The first line of the input contains a pair of integers n, m (1 ≤ n ≤ 3000, 0 ≤ m ≤ 30000) — the number of intersections and roads, respectively. Next m lines list the roads, one per line. Each of the roads is given by a pair of integers a_i, b_i (1 ≤ a_i, b_i ≤ n;a_i ≠ b_i) — the number of the intersection it goes out from and the number of the intersection it leads to. Between a pair of intersections there is at most one road in each of the two directions.

It is not guaranteed that you can get from any intersection to any other one.

Output

Print the required number of "damn rhombi".

Sample test(s)

Input

Output

Input

Output

题意：问一个给定的有向图中有几个如图所示的四边形。

思路：对每个点进行两层BFS，并记录第二层的每个点被访问的次数，注意第二层的点一定不能是起点！然后就可知道每一个第二层点和起点构成的四边行的个数。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
struct EDGE
{
    int to,next;
}edge[30005];
int edgecnt,head[3005];
void init()
{
    edgecnt=0;
    memset(head,-1,sizeof(head));
}
inline void add(int s,int t)
{
    edge[edgecnt]=(EDGE){t,head[s]};
    head[s]=edgecnt++;
}
int n;
int ans;
void bfs(int x)
{
    int vis[3005];
    memset(vis,0,sizeof(vis));
    queue<int>q;
    for(int i=head[x];~i;i=edge[i].next) q.push(edge[i].to);
    while(!q.empty())
    {
        int t=q.front();
        q.pop();
        for(int i=head[t];~i;i=edge[i].next)
        if(edge[i].to!=x) vis[edge[i].to]++;
    }
    for(int i=1;i<=n;i++) if(vis[i]>=2) ans+=vis[i]*(vis[i]-1)/2;
}
int main()
{
    int m;
    scanf("%d %d",&n,&m);
    ans=0;
    init();
    while(m--)
    {
        int s,t;
        scanf("%d %d",&s,&t);
        add(s,t);
    }
    for(int i=1;i<=n;i++) bfs(i);
    printf("%d\n",ans);
    return 0;
}