BNUOJ 1585 Girls and Boys

Girls and Boys

Time Limit: 5000ms

Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1466
64-bit integer IO format: %lld      Java class name: Main

 

 

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

解题：找出相互没关系的一个人数最大的集合。首先，是点！让点多，我们可以求点小！最小点覆盖！那么我们可以求最大匹配数！无向图的最大匹配数要处以2.。。。就这样了

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 struct arc{
     int to,next;
 };
 arc g[];
 int head[maxn],from[maxn],tot,n;
 bool vis[maxn];
 void add(int u,int v){
     g[tot].to = v;
     g[tot].next = head[u];
     head[u] = tot++;
 }
 bool dfs(int u){
     for(int i = head[u]; i != -; i = g[i].next){
         if(!vis[g[i].to]){
             vis[g[i].to] = true;
             if(from[g[i].to] == - || dfs(from[g[i].to])){
                 from[g[i].to] = u;
                 return true;
             }
         }

     }
     return false;
 }
 int main() {
     int i,j,k,u,v,ans;
     while(~scanf("%d",&n)){
         memset(head,-,sizeof(head));
         memset(from,-,sizeof(from));
         tot = ;
         for(i = ; i < n; i++){
             scanf("%d: (%d)",&j,&k);
             for(j = ; j < k; j++){
                 scanf("%d",&v);
                 add(i,v);
                 add(v,i);
             }
         }
         for(ans = i = ; i < n; i++){
             memset(vis,false,sizeof(vis));
             if(dfs(i)) ans++;
         }
         ans >>= ;
         printf("%d\n",n-ans);
     }
     return ;
 }