HDU - 5952 Counting Cliques(DFS)

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.

InputThe 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.OutputFor 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

思路：如何找到一个k阶的完全图？如果一个图是完全图，那么引入一个新的点，这个点与原图中的每一点都有边相连，新图还是完全图。用了num数组来记录原图上的点。建图时，、只建了从编号小的点到编号大的点之间的边。这是由于，每次没有必要建立反向边。反而不建反向边的话，会少了去重的过程。代码：

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 108;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
vector<int>u[maxn];
int n,m,k;
int ans;
int top;
int num[maxn];
bool mp[maxn][maxn];
void dfs(int t,int d)
{
    if(d==k){ans++;return;}
    int siz = u[t].size();
    bool flag = false;
    for(int i=0;i<siz;i++){
        int cnt = u[t][i];
        flag = false;
        for(int j=1;j<=top;j++){
            if(!mp[cnt][num[j]]){flag = true;break;}
        }
        if(flag){continue;}

        num[++top]=cnt;
        dfs(cnt,d+1);
        top--;
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        ans = 0;
        scanf("%d%d%d",&n,&m,&k);
        memset(mp,0,sizeof(mp));
        for(int i=1;i<=n;i++){
            u[i].clear();
        }
        int x,y;
        for(int i=1;i<=m;i++){
            scanf("%d%d",&x,&y);
            if(x>y){swap(x,y);}
            u[x].push_back(y);
            mp[x][y]=mp[y][x]=true;
        }

        for(int i=1;i<=n;i++){
            num[++top]=i;
            dfs(i,1);
            top--;
        }
        printf("%d\n",ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/ZGQblogs/p/9890878.html