Girls and Boys

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9245    Accepted Submission(s): 4240

Problem Description

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.

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

题解：男生女生可能会相互吸引，就爱上了彼此，所以让你写一个程序，让找出不让他们相互喜欢的集合数目；

代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<vector>
 7 #define mem(x,y) memset(x,y,sizeof(x))
 8 using namespace std;
 9 const int INF=0x3f3f3f3f;
10 const int MAXN=1010;
11 vector<int>vec[MAXN];
12 int usd[MAXN],vis[MAXN];
13 bool dfs(int x){
14     for(int i=0;i<vec[x].size();i++){
15         int v=vec[x][i];
16             if(!vis[v]){
17                 vis[v]=1;
18             if(usd[v]==-1||dfs(usd[v])){
19                 usd[v]=x;return true;
20             }
21         }
22     }
23     return false;
24 }
25 int main(){
26     int N,a,t;
27     while(~scanf("%d",&N)){
28         for(int i=0;i<MAXN;i++)vec[i].clear();
29         for(int i=0;i<N;i++){
30             scanf("%*d: (%d)",&t);
31         //    printf("t=%d\n",t);
32             while(t--){
33                 scanf("%d",&a);
34                 vec[i].push_back(a);
35             }
36         }mem(usd,-1);mem(vis,0);
37         int ans=0;
38         for(int i=0;i<N;i++){
39             mem(vis,0);
40             if(dfs(i))ans++;
41         }
42         printf("%d\n",N-ans/2);
43     }
44     return 0;
45 }