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.
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
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
Hungarian algorithm is proposed by the Hungarian mathematician Edmonds in 1965, hence the name. Hungarian algorithm is thought sufficient proof of the theorem Hall-based, it is the most common of the Ministry of graph matching algorithm, the core of the algorithm is to find augmenting path, it is a demand by augmenting path bipartite graph maximum matching algorithm.
------- etc., see big head? So consider the following version:
Note: The following switched http://blog.csdn.net/dark_scope/article/details/8880547
By generations of effort, you finally caught up with the tide of left men left woman, assuming you are a glorious new century matchmaker, there are N remaining men in your hands, M a woman left, everyone is likely to more than Interested in heterosexual ( 
-_- || not to consider specific sexual orientation), if a couple each have a crush, then you can put the pair together, to now let's ignore all out of unrequited love (good sad feeling 
), you have probably follows a diagram, each connection have expressed mutual goodwill.
The spirit of a life, build a seven wins Buddha principle of saving lives, the more you try to fix a couple as much as possible, working mode Hungary algorithm will teach you to do this:
===============================================================================
A: first try to find the sister No. 1 boys, girls first discover No. 1 and his name is connected also to spend no master, got it, even on a blue line
===============================================================================
Two : Next to the No. 2 guy looking for my sister, Discovery spent the first 2 girls name and he connected without the Lord, got it
===============================================================================
Three : Next is the No. 3 boys, we regret No. 1 girl already has a primary, and how to do it?
We tried to No. 1 before the boys and girls matches (that is, No. 1 boys) additionally assigned a sister.
(Yellow indicates that the edge is temporarily removed)
No. 1 boys and girls is connected to the second number 2 girls, 2 girls but also the Lord, how to do it? We then try to wife No. 2 girls ( 
) to re-find a sister (Note that this step is the same as above, which is a recursive process)
At this time, the boys found No. 2 No. 3'll find the girls, then the problem is solved before, back to back
No. 2 No. 3 boys can find sister ~ ~ ~ No. 1 boys can find the sister No. 2 ~ 3 boys can find a number 1 sister
The third step so the final result is:
===============================================================================
四： 接下来是4号男生，很遗憾，按照第三步的节奏我们没法给4号男生腾出来一个妹子，我们实在是无能为力了……香吉士同学走好。
===============================================================================
其原则大概是：有机会上，没机会创造机会也要上.
这是匈牙利算法的理解；
然后再看这道题；
#include <stdio.h>
#include <string.h>
int book[10010];
int match[10010];
int y[1010][1010];
int n;
int yue(int u) //匈牙利算法
{
int v,i,j;
for(i=1; i <= n; i++){
if(book[i] == 0 && y[u][i] == 1){
book[i]=1;
if(match[i]==0||yue(match[i])){
match[i]=u;
return 1;
}
}
}
return 0;
}
int main()
{
int i,j,k,a,b,m;
while(scanf("%d",&n)!=EOF)
{
memset(y,0,sizeof(y));
memset(match,0,sizeof(match));
for(i=0;i<n;i++){
scanf("%d: (%d)",&a,&m);
the while (m -) {
Scanf ( "% D", & B);
Y [A +. 1] [B +. 1] =. 1;
}
}
K = 0;
for (I =. 1; I <= n-; I ++) {
Memset (Book, 0, the sizeof (Book));
IF (Yue (I) ==. 1)
K ++;
}
the printf ( "% D \ n-", NK / 2); // since this question (1, 2) with (2,1) is the same, it is divided by k to obtain 2;
}
return 0;
}