1530 Maximum Clique 最大团（模板）

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Maximum Clique

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2107    Accepted Submission(s): 1117

Problem Description

Given a graph G(V, E), a clique is a sub-graph g(v, e), so that for all vertex pairs v1, v2 in v, there exists an edge (v1, v2) in e. Maximum clique is the clique that has maximum number of vertex.

 

 

Input

Input contains multiple tests. For each test:

The first line has one integer n, the number of vertex. (1 < n <= 50)

The following n lines has n 0 or 1 each, indicating whether an edge exists between i (line number) and j (column number).

A test with n = 0 signals the end of input. This test should not be processed.

 

 

Output

One number for each test, the number of vertex in maximum clique.

 

 

Sample Input

50 1 1 0 11 0 1 1 11 1 0 1 10 1 1 0 11 1 1 1 00

 

 

Sample Output

4

 

 

Author

CHENG, Long

 

 

Source

ZOJ Monthly, February 2003

 

 

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最大团的模板题。详细解释：点击打开链接。

#include<stdio.h>#define M 51int g[M][M],cs[M],ans[M],now[M];int re;bool flag;//最大团模板//////////////////////////////void dfs(int n,int depth,int *u){    int i,j,k,v[M],vn;    if(n)    {        if(depth+cs[u[0]]<=re)            return;        for(i=0;i<n+depth-re;i++)        {            for(j=i+1,vn=0;j<n;j++)                if(g[u[i]][u[j]])                    v[vn++]=u[j];            now[depth]=u[i];            dfs(vn,depth+1,v);            if(flag)                return;        }    }    else if(depth>re)    {        re=depth;        flag=true;        for(i=0;i<depth;i++)            ans[i]=now[i];    }}int clique(int n){    int vn,v[M],i,j,k;    re=0;    for(cs[i=n-1];i>=0;i--)    {        for(vn=0,j=i+1;j<n;j++)            if(g[i][j])                v[vn++]=j;        flag=false;        now[0]=i;        dfs(vn,1,v);        cs[i]=re;    }    return re;}//////////////////////////////////////int main(){    int n;    while(scanf("%d",&n),n)    {        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                scanf("%d",&g[i][j]);        printf("%d\n",clique(n));    }    return 0;}