【HDU 5952 Counting Cliques】& DFS_acm 搜索大小为4的团的数量-优快云博客

该博客讨论了如何解决图论问题，即在给定的图中计数具有特定大小S的团（完全子图）。通过深度优先搜索（DFS）算法，逐个检查节点并确保每个加入的节点与现有节点之间存在边，以确保形成的子图是完全连通的。文章提供了样例输入和输出，并给出了实现解决方案的AC代码。

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Counting Cliques

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2715 Accepted Submission(s): 982

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

Source
2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

题意：给出的图中有多少含 S 个节点的完全图

思路 : DFS 搜一下，每加入一个点便判断这个点和其他点是否有边，保证加入的这个点后依然是完全图，为避免重复搜索没必要建双向边

AC代码：

#include<cstdio>
#include<cmath>
#include<deque>
#include<queue>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e5 + 10;
const int INF = 1e9 + 7;
typedef long long LL;
int n,m,s,nl,head[MAX * 2],d[110][110];
int ans,u[110];
struct node{
    int to,next;
}st[MAX * 2];
void add(int a,int b){
    st[nl].to = b;
    st[nl].next = head[a];
    head[a] = nl++;
}
void dfs(int a,int b){
    if(b == s){
        ans++;
        return ;
    }
    for(int i = head[a]; i != -1; i = st[i].next){
        int o = st[i].to,ok = 1;
        for(int j = 1; j <= b; j++)
            if(!d[o][u[j]]){
                ok = 0;
                break;
            }
        if(ok){
            u[b + 1] = o;
            dfs(o,b + 1);
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        nl = 0;
        memset(head,-1,sizeof head);
        scanf("%d %d %d",&n,&m,&s);
        memset(d,0,sizeof d);
        for(int i = 1; i <= m; i++){
            int a,b;
            scanf("%d %d",&a,&b);
            d[a][b] = d[b][a] = 1;
            add(a,b);
        }
        if(s == 2)
            printf("%d\n",m);
        else{
            ans = 0;
            for(int i = 1; i <= n; i++)
                u[1] = i,dfs(i,1);
            printf("%d\n",ans);
        }
    }
    return 0;
}