hdu 6184 Counting Stars（求无向图三元环）

Bertha 。 2022-06-09 12:26 130阅读 0赞

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) V=(A,B,C,D) and E=(AB,BC,CD,DA,AC) E=(AB,BC,CD,DA,AC), we call G as an "A-structure".  
  
It is defined that "A-structure" G1=V1\+E1 G1=V1+E1 and G2=V2\+E2 G2=V2+E2 are same only in the condition that V1=V2 V1=V2 and E1=E2 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 2≤n≤105, 1≤m≤min(2×105,n(n−1)2) 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 1≤u,v≤n  
  
∑n≤3×105 ∑n≤3×105, ∑m≤6×105 ∑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

题解：

比赛的时候完全不会做。。。后来搜了个大佬的博客：[http://blog.csdn.net/jaihk662/article/details/77750832][http_blog.csdn.net_jaihk662_article_details_77750832]

看懂了几乎照着打了一遍。。。不过我觉得这里面用map会比用set更好理解

做法就是定义一个t为sqrt(m)，遍历每一个点，先遍历一次它所有连接的边，将相邻的点打上link标记，然后再遍历一遍，看另一个点的度数是否大于t，小于就用o（n^3）暴力做法遍历该点的所有边看是否可以连回初始点，否则的话就枚举初始点的所有连接的点是否连通。。。然后这里还要用一个组合数学，题目要求的是要两两三元环组成的四元环的个数。。就是每一个点的三元环数目sum\*（sum-1）/2，累加就是结果

代码：

#include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<vector>
    #include<deque>
    #include<algorithm>
    using namespace std;
    #define INF 100861111
    #define ll long long
    #define eps 1e-7
    #define lson k*2
    #define rson k*2+1
    const int maxn=100005;
    vector<int>p[maxn];
    map<pair<int,int>,int>M;
    int vis[maxn],link[maxn],out[maxn];
    int main()
    {
        ll ans,sum;
        int n,i,j,m,k,x,y,z,t;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            t=sqrt(m);
            M.clear();
            for(i=1;i<=n;i++)
            {
                p[i].clear();
                vis[i]=out[i]=link[i]=0;
            }
            for(i=1;i<=m;i++)
            {
                scanf("%d%d",&x,&y);
                p[x].push_back(y);
                p[y].push_back(x);
                out[x]++;
                out[y]++;
                pair<int,int>temp1=make_pair(x,y);
                M[temp1]=1;
                pair<int,int>temp2=make_pair(y,x);
                M[temp2]=1;
            }
            ans=0;
            for(i=1;i<=n;i++)
            {
                x=i;
                vis[x]=1;
                for(j=0;j<p[x].size();j++)
                    link[p[x][j]]=x;
                for(j=0;j<p[x].size();j++)
                {
                    sum=0;
                    y=p[x][j];
                    if(vis[y])
                        continue;
                    if(out[y]<=t)
                    {
                        for(k=0;k<p[y].size();k++)
                        {
                            z=p[y][k];
                            if(link[z]==x)
                                sum++;
                        }
                    }
                    else
                    {
                        for(k=0;k<p[x].size();k++)
                        {
                            z=p[x][k];
                            pair<int,int>temp=make_pair(z,y);
                            if(M[temp]!=0)
                                sum++;
                        }
                    }
                    ans+=sum*(sum-1)/2;
                }
            }
            printf("%lld\n",ans);
        }
        return 0;
    }

[http_blog.csdn.net_jaihk662_article_details_77750832]: http://blog.csdn.net/jaihk662/article/details/77750832