HDU5952Counting Cliques(dfs)

Counting Cliques
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1170 Accepted Submission(s): 449

Problem Description

A clique is a complete graph, in which there is an edge between every pair of the vertices. Given a graph with N vertices and M edges, your task is to count the number of cliques with a specific size S in the graph.

Input

The first line is the number of test cases. For each test case, the first line contains 3 integers N,M and S (N ≤ 100,M ≤ 1000,2 ≤ S ≤ 10), each of the following M lines contains 2 integers u and v (1 ≤ u < v ≤ N), which means there is an edge between vertices u and v. It is guaranteed that the maximum degree of the vertices is no larger than 20.

Output

For each test case, output the number of cliques with size S in the graph.

Sample Input

3
4 3 2
1 2
2 3
3 4
5 9 3
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
6 15 4
1 2
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6
4 5
4 6
5 6

Sample Output

3
7
15
题意：给你n个点m条边的无向图，问顶点数为s的团的个数。
题解：dfs 枚举每一个点，看他能形成多少个满足条件的团，然后剩下的团不会含有上一个点，保证团不会重复计算。
代码：

#include<stdio.h>#include<string.h>#include<iostream>#include <map>#include <vector>using namespace std;const int N=1200;vector<int>ed[N];int mp[N][N];int n,m,s,t,ans,x,y;void dfs(int u,int *cnt,int size){    if(size==s)    {        ans++;        return ;    }    bool p;    for(int i=0; i<ed[u].size(); i++)    {        int v=ed[u][i];        p=true;        for(int j=1; j<=size; j++)        {            if(!mp[cnt[j]][v])            {                p=false;                break;            }        }        if(p)        {            size++;            cnt[size]=v;            dfs(v,cnt,size);            cnt[size]=0;            size--;        }    }}void init(){    for(int i=1; i<=n; i++)    {        ed[i].clear();    }    memset(mp,0,sizeof(mp));    ans=0;}int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&n,&m,&s);        init();        for(int i=0; i<m; i++)        {            scanf("%d%d",&x,&y);            if(x>y)                swap(x,y);            ed[x].push_back(y);            mp[x][y]=mp[y][x]=1;        }        for(int i=1; i<=n; i++)        {            int size=1;            int cnt[N];            cnt[1]=i;            dfs(i,cnt,size);        }        printf("%d\n",ans);    }}