POJ 1466 Girls and Boys (匈牙利算法 最大独立集)

Girls and Boys

Time Limit: 5000MS Memory Limit: 10000KTotal Submissions: 10912 Accepted: 4887

Description

In 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.

Input

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 (n <=500 ), for n subjects.

Output

For each given data set, the program should write to standard output a line containing the result.

Sample Input

70: (3) 4 5 61: (2) 4 62: (0)3: (0)4: (2) 0 15: (1) 06: (2) 0 130: (2) 1 21: (1) 02: (1) 0

Sample Output

52

Source

Southeastern Europe 2000

题目链接：http://poj.org/problem?id=1466

题目大意：一些男的和一些女的有的之间有关系，同性之间无关系，求一个最大的集合，使得里面任意两人没有关系

题目分析：裸的最大独立集问题，建立二分图，然后根据最大独立集＝点数－最大匹配，求解即可，注意由于二分图是双向建立的，求出的最大匹配要除2

#include <cstdio>#include <cstring>int const MAX = 505;bool g[MAX][MAX];bool vis[MAX];int cx[MAX], cy[MAX];int n;int DFS(int x){    for(int y = 0; y < n; y++)    {        if(!vis[y] && g[x][y])        {            vis[y] = true;            if(cy[y] == -1 || DFS(cy[y]))            {                cy[y] = x;                cx[x] = y;                return 1;            }        }    }    return 0;}int MaxMatch(){    memset(cx, -1, sizeof(cx));    memset(cy, -1, sizeof(cy));    int res = 0;    for(int i = 0; i < n; i++)    {        if(cx[i] == -1)        {            memset(vis, false, sizeof(vis));            res += DFS(i);        }    }    return res;}int main(){    while(scanf("%d", &n) != EOF)    {        memset(g, false, sizeof(g));        for(int i = 0; i < n; i++)        {            int x;            scanf("%d", &x);            int num;            scanf(": (%d)", &num);            for(int j = 0; j < num; j++)            {                int y;                scanf("%d", &y);                g[x][y] = true;            }        }        int ans = MaxMatch();        printf("%d\n", n - ans / 2);    }}