ZOJ1137(Girls and Boys)

Girls and Boys

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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the second year of the university somebody started a study on the romantic relations between the students. The relation ��romantically involved�� is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been ��romantically involved��. The result of the program is the number of students in such a set.

The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:

the number of students
the description of each student, in the following format
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
or
student_identifier:(0)

The student_identifier is an integer number between 0 and n-1, for n subjects.
For each given data set, the program should write to standard output a line containing the result.

An example is given in Figure 1.

Input

7
0: (3) 4 5 6
1: (2) 4 6
2: (0)
3: (0)
4: (2) 0 1
5: (1) 0
6: (2) 0 1
3
0: (2) 1 2
1: (1) 0
2: (1) 0

Output

5
2

//1860068 2009-05-08 15:57:14 Accepted  1137 C++ 2820 1172 aslys 
/**//* 

    ZOJ1137 Girls and Boys
        by Xredman
*/
#include <iostream>
using namespace std;

const int N = 1000;//the number of students

bool mat[N][N];
int useif[N];
int cx[N], cy[N];
int n;

bool dfs(int cc)
{
    int i;
    for(i = 0; i < n; i++)
    {
        if(!useif[i] && mat[cc][i])
        {
            useif[i] = true;
            if(cy[i] == -1 || dfs(cy[i]))
            {
                cx[cc] = i;
                cy[i] = cc;
                return true;
            }
        }
    }
    return false;
}

int MaxMatch()
{//最大二分匹配，匈牙利实现
    int i, sum = 0;
    memset(cx, -1, sizeof(cx));
    memset(cy, -1, sizeof(cy));

    for(i = 0; i < n; i++)
    {
        memset(useif, false, sizeof(useif));
        if(dfs(i))
            sum++;
    }
    return sum;
}

int main()
{
    int st, x, y;
    int i;

    while(scanf("%d", &n) != EOF)
    {
        memset(mat, false, sizeof(mat));
        for(i = 0; i < n; i++)
        {//init
            scanf("%d: (%d)", &x, &st);
            while(st--)
            {
                scanf("%d", &y);

                mat[x][y] = true;
            }
        }
        cout<<n - MaxMatch() / 2<<endl;
    }
    return 0;
}
//0: (3) 4 5 6