hdu 1530 Maximum Clique（最大团模版）

Maximum Clique

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2111    Accepted Submission(s): 1118

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
题意：无向图的最大团问题。
思路：回溯法，详见http://www.cnblogs.com/pushing-my-way/archive/2012/08/08/2627993.html
AC代码：
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <vector>#include <cmath>#include <cstdlib>#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)#define ll long long#define eps 1e-6using namespace std;const int maxn=55;const int INF=1000000000;int n,cnt,bestn;int G[maxn][maxn];int vis[maxn];void dfs(int u){    if(u>n)    {        if(cnt>bestn) bestn=cnt;        return;    }    bool ok=true;    for(int i=0;i<u;i++)    if(vis[i]&&!G[i][u])    {        ok=false;        break;    }    if(ok)    {        vis[u]=true;        cnt++;        dfs(u+1);        vis[u]=false;        cnt--;    }    if(cnt+n-u>bestn) dfs(u+1);}int main(){    while(scanf("%d",&n),n)    {        for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)        scanf("%d",&G[i][j]);        cnt=bestn=0;        memset(vis,false,sizeof(vis));        dfs(1);        printf("%d\n",bestn);    }}