1466 Girls and Boys //MaxMatch

最新推荐文章于 2022-04-19 22:03:08 发布

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本文介绍了一个基于最大匹配算法的问题——如何找出大学里未建立浪漫关系的最大学生集合。通过定义浪漫关系为一名男生与一名女生之间的关系，文章提供了一种解决方法，使用邻接矩阵和匈牙利算法的变体来找到最大匹配数，从而计算出最大独立集的学生数量。

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Girls and Boys

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 5324		Accepted: 2304

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

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

Sample Output

5
2

Source

Southeastern Europe 2000

别忘了除以2

因为男女生的人数没有规定，所以有重复计算

所以除以2，就把重复计算给滤掉了。

#include<stdio.h>

#include<string.h>

bool mat[505][505];

bool usedif[505];

int link[505];

int n;

bool can(int t)

{

int i;

for(i=0;i<n;i++)

if(!usedif[i]&&mat[t][i])

{

usedif[i]=true;

if(link[i]==-1||can(link[i]))

{

link[i]=t;

return true;

}

return false;

}

int maxMatch()

{

int i,num=0;

memset(link,-1,sizeof(link));

for(i=0;i<n;i++)

{

memset(usedif,0,sizeof(usedif));

if(can(i))

num++;

}

return num;

}

int main()

{

int i,u,v,m,ans;

while(scanf("%d",&n)!=EOF)

{

if(n==0)

break;

memset(mat,0,sizeof(mat));

for(i=0;i<n;i++)

{

scanf("%d: (%d)",&u,&m); //注意 //注意这里的输入

while(m--)

{

scanf("%d",&v);

mat[u][v]=true;

}

ans=n-maxMatch()/2;

printf("%d/n",ans);

}

return 0;

}