hdu 1530 Maximum Clique (最大包)

Maximum Clique
Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5380    Accepted Submission(s): 2776

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
5
0 1 1 0 1
1 0 1 1 1
1 1 0 1 1
0 1 1 0 1
1 1 1 1 0
0
 

Sample Output
4

 

C/C++：

#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0xffffff
using namespace std;
const int my_max = 55;

int n, my_map[my_max][my_max], my_ans, my_set[my_max];

bool my_is_clique(int my_depth, int my_point)
{
    for (int i = 1; i < my_depth; ++ i)
        if (!my_map[my_set[i]][my_point])
            return false;
    return true;
}

void my_dfs(int my_depth, int my_point)
{
    //if (my_depth + n - my_point + 1 <= my_ans) return;
    for (int i = my_point; i <= n; ++ i)
    {
        if (my_is_clique(my_depth + 1, i))
        {
            my_set[my_depth + 1] = i;
            my_dfs(my_depth + 1, i + 1);
        }
    }
    if (my_depth > my_ans) my_ans = my_depth;
}

int main()
{
    while (~scanf("%d", &n), n)
    {
        memset(my_map, 0, sizeof(my_map));
        memset(my_set, 0, sizeof(my_set));

        for (int i = 1; i <= n; ++ i)
            for (int j = 1; j <= n; ++ j)
                scanf("%d", &my_map[i][j]);
        my_ans = 0;
        my_dfs(0, 1);
        printf("%d\n", my_ans);
    }
    return 0;
}