hdu 5952Counting Cliques(暴搜)

题目链接

Counting Cliques

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1371    Accepted Submission(s): 527

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

34 3 21 22 33 45 9 31 31 41 52 32 42 53 43 54 56 15 41 21 31 41 51 62 32 42 52 63 43 53 64 54 65 6

 

Sample Output

3715

 

题解：

因为数据量比较小，直接爆搜就可以了，不过有一点要注意，爆搜时用一个数组记录在完全图里面的点，而且为了保证不重复，这些点按照递增排列，那么对于边u-v(u<v)此时图就只需要存u->v就行了，不然会T。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>#include<queue>#include<stack>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const int inf=0x3fffffff;const ll mod=1000000007;const int maxn=100+10;int g[maxn][maxn];int head[maxn];struct edge{    int from,to,next;}e[maxn*20];   //int tol=0;void add(int u,int v){    e[++tol].to=v,e[tol].next=head[u],head[u]=tol;}void init(){    memset(head,0,sizeof(head));    memset(g,0,sizeof(g));    tol=0;}int ans;int n,m,s;void dfs(int u,int sz,int *tmp){    if(sz==s)    {        ans++;        return;    }    for(int i=head[u];i;i=e[i].next)    {        int v=e[i].to;        //if(v<u) continue;        bool flag=true;        for(int j=1;j<=sz;j++)        {            if(!g[tmp[j]][v])            {                flag=false;                break;            }        }        if(flag)        {            tmp[sz+1]=v;            dfs(v,sz+1,tmp);        }    }    }int main(){    int cas;    scanf("%d",&cas);    while(cas--)    {        scanf("%d%d%d",&n,&m,&s);        init();        while(m--)        {            int a,b;            scanf("%d%d",&a,&b);            g[a][b]=g[b][a]=1;            if(a>b) swap(a,b);            add(a,b);    //只保存每个点比他大的相邻点            //add(b,a);        }        ans=0;        for(int i=1;i<=n;i++)        {            int tmp[maxn];            memset(tmp,0,sizeof(tmp));            tmp[1]=i;            dfs(i,1,tmp);        }        printf("%d\n",ans);    }    return 0;}