Counting Cliques

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3480 Accepted Submission(s): 1254

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亞洲區瀋陽站-重現賽（感謝東北大學）

Recommend
jiangzijing2015

題目大意

　　有一張最多100個點的無向圖，求大小為S的團有多少個。

解題思路

　　最大團極大團問題是NP問題，所以這題也很自然地想到搜尋。不過如果直接暴搜的話可以發現對於每個大小為S的團會被搜到S次。考慮怎麼優化這裡，考慮對於任一一個團一定有且僅有一條按照頂點序號走過所有點的路徑。所以我們就可以對於每條邊定向為從編號小的點到編號大的點。這樣再暴搜的時候就優化了非常多，就可以通過這道題了。

AC程式碼

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f

const int MAXV=100+3;

int V, E, S;
vector<int> G[MAXV];
bool maze[MAXV][MAXV];
int ans;
int save[MAXV];

void init()
{
    for(int i=0;i<=V;++i)
        G[i].clear();
    for(int i=0;i<=V;++i)
        for(int j=0;j<=V;++j)
            maze[i][j]=false;
    ans=0;
}

void dfs(int u, int n)
{
    if(n==S)
    {
        ++ans;
        return ;
    }
    for(int i=0;i<G[u].size();++i)
    {
        int v=G[u][i];
        bool ok=true;
        for(int j=0;j<n;++j)
            if(!maze[save[j]][v])
            {
                ok=false;
                break;
            }
        if(!ok)
            continue;
        save[n]=v;
        dfs(v, n+1);
    }
}

int main()
{
    int T_T;
    scanf("%d", &T_T);
    while(T_T--)
    {
        scanf("%d%d%d", &V, &E, &S);
        init();
        for(int i=0;i<E;++i)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            if(u<=v)
            {
                G[u].push_back(v);
                maze[u][v]=true;
            }
            else
            {
                G[v].push_back(u);
                maze[v][u]=true;
            }
        }
        for(int u=0;u<V;++u)
            if(G[u].size()+1>=S)
            {
                save[0]=u;
                dfs(u, 1);
            }
        printf("%d\n", ans);
    }

    return 0;
}