HDU-5952 Counting Cliques(16年ICPC沈阳赛区)(暴力DFS)

Counting Cliques

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 673    Accepted Submission(s): 264


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
For each test case, output the number of cliques with size S in the graph.
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

3
7
15

 

[align=left]Source[/align]
2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学） 

题意:给出一个无向图,要求求出点的个数恰好为s的完全子图的数目

题解:(万能的)暴力,具体原理看代码吧= =

#include<cstdio>
#include<algorithm>
#include<string.h>
#include<bitset>
using namespace std;
const int maxn = 105;
int s,ans;
bool mp[maxn][maxn];
/*
这里的dfs看起来很简单,其实包含了一个很大的剪枝:在遍历某个点的时候,已经确定了已经加入团的点是与这个点相连的,还可以知道平均下来每个点的入度不会超过20,
因此整个遍历的复杂度最高是C(9,20)*100
*/
void dfs(int mx,int v[],int cnt){
int nxt[maxn];
if(cnt==s) {ans++;return;}
for(int i=0;i<mx;i++){           //枚举能与点数为cnt的团构成点数为cnt+1的团的所有点
int len=0;
for(int j=i+1;j<mx;j++){
if(mp[v[i]][v[j]]){    //该团的所有"候选点"必须与已经选了的点相连
nxt[len++]=v[j];
}
}
dfs(len,nxt,cnt+1);
}
}
int main(){
int T,n,m,u,v;
// freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&n,&m,&s);
memset(mp,0,sizeof(mp));
for(int i=0;i<m;i++){
scanf("%d%d",&u,&v);
mp[u][v]=mp[v][u]=1;
}
ans=0;
int nxt[105];
for(int i=1;i<=n;i++){      //枚举每个点,这个点必须在所求的团里面,以这个点开始搜索
int len=0;
for(int j=i+1;j<=n;j++){
if(mp[i][j]){      //该团的所有"候选点"必须与已经选了的点相连
nxt[len++]=j;
}
}
dfs(len,nxt,1);
}
printf("%d\n",ans);
}
return 0;
}