【2017广西邀请赛】hdu 6184 Counting Stars 三元环+set

Problem Description

Little A is an astronomy lover, and he has found that the sky was so beautiful!

So he is counting stars now!

There are n stars in the sky, and little A has connected them by m non-directional edges.

It is guranteed that no edges connect one star with itself, and every two edges connect different pairs of stars.

Now little A wants to know that how many different "A-Structure"s are there in the sky, can you help him?

An "A-structure" can be seen as a non-directional subgraph G, with a set of four nodes V and a set of five edges E.

If V=(A,B,C,D) and E=(AB,BC,CD,DA,AC),
we call G as an "A-structure".

It is defined that "A-structure" G1=V1+E1 and G2=V2+E2 are
same only in the condition that V1=V2 and E1=E2.

 

Input

There are no more than 300 test cases.

For each test case, there are 2 positive integers n and m in the first line.

2≤n≤105, 1≤m≤min(2×105,n(n−1)2)

And then m lines follow, in each line there are two positive integers u and v, describing that this edge connects node u and node v.

1≤u,v≤n

∑n≤3×105,∑m≤6×105

 

Output

For each test case, just output one integer--the number of different "A-structure"s in one line.

 

Sample Input

4 5
1 2
2 3
3 4
4 1
1 3
4 6
1 2
2 3
3 4
4 1
1 3
2 4

 

Sample Output

1
6

 

题意：

n个点m条边的无向图，问有多少个A-structure

其中A-structure满足V=(A,B,C,D) && E=(AB,BC,CD,DA,AC)

思路：

这东西就是两个有公共边的三元环。所以问题可以退化成求三元环，有个暴力的msqrtm的算法。

遍历每个点x，其相邻的点为y，如果y的度小于sqrt(m)，那遍历y相邻的点z，判断x与z是否相连。因为遍历x和y相当于遍历m条边，而y相邻的点不超过sqrt(m)，所以复杂度msqrtm；

如果y的度大于sqrt(m)，那遍历x相邻的点z，判断y和z是否相连。仔细思考可以看出，这里相当于遍历了所有边的基础上遍历度大于sqrt(m)的点，在因为度大于sqrt(m)的点个数不超过sqrt(m)，所以这里的复杂度为msqrtm

判断相连可以用多种方式，最简单的就是map，但这题不能用map，会超内存。用set 的话会超时，但是可以在第一种情况下可以用数组判断来优化。

//
// main.cpp
// 1003
//
// Created by zc on 2017/9/8.
// Copyright © 2017年 zc. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#define ll long long
using namespace std;
const int N=100005;
int n,m,x,y,du
,fa
;
bool v
;
set<ll>st;
vector<int>r
;

int main(int argc, const char * argv[]) {
while(~scanf("%d%d",&n,&m))
{
st.clear();
for(int i=1;i<=n;i++) r[i].clear(),fa[i]=v[i]=du[i]=0;
for(int i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
r[x].push_back(y);
r[y].push_back(x);
du[x]++;du[y]++;
st.insert((ll)x*n+y);
st.insert((ll)y*n+x);
}
memset(v,0,sizeof(v));
ll ans=0;
for(int i=1;i<=n;i++)
{
v[i]=1;
for(int j=0;j<r[i].size();j++) fa[r[i][j]]=i;
for(int j=0;j<r[i].size();j++)
{
int u=r[i][j];
if(v[u]) continue;
int sum=0;
if(du[u]<=(int)sqrt(m))
{
for(int k=0;k<r[u].size();k++)
if(fa[r[u][k]]==i) sum++;
}
else
{
for(int k=0;k<r[i].size();k++)
if(st.find((ll)u*n+r[i][k])!=st.end()) sum++;
}
ans+=(ll)sum*(sum-1)/2;
}
}
printf("%lld\n",ans);
}
}