Girls and Boys

Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 249 Accepted Submission(s): 163

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.

 

Output

            

 

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

 

Recommend

JGShining

 

/*
初次思考：给出你一个点阵，然后让你求最大连通图的点数
并查集想法，找点

#错误：题意没理解明白

题意：找出一个最大集合使得任意两个人没有关系
因为有关系的肯定是男女（暂时不考虑性取向的问题）所以就将整个关系图，联系到了二分图上，进而用到二分匹配问题的解决方案
用到“匈牙利算法”

要求最大的没有关系的集合，就是总顶点数-最大匹配数
*/
#include<bits/stdc++.h>
using namespace std;
int n,u,m,v;
/***********************二分匹配模板**************************/
const int MAXN=1010;
int g[MAXN][MAXN];//编号是0~n-1的 
int linker[MAXN];//记录匹配点i的匹配点是谁
bool used[MAXN];
bool dfs(int u)//回溯看能不能通过分手来进行匹配
{
    int v;
    for(v=0;v<n;v++)
        if(g[u][v]&&!used[v])
        //如果有这条边，并且这条边没有用过
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))//如果这个点没有匹配过，并且能找到匹配点，那么就可以以这个边作为匹配点
            {
                linker[v]=u;
                return true;
            }    
        }  
    return false;  
}    
int hungary()//返回最大匹配数
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=0;u<n;u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))//如果这个点有匹配点 
            res++;
    } 
    return res;   
}
/***********************二分匹配模板**************************/
void init(){
    memset(g,0,sizeof g);
}
int main(){
    //freopen("C:\Users\acer\Desktop\in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        init();
        for(int i=0;i<n;i++){
            scanf("%d: (%d)",&u,&m);
            for(int j=0;j<m;j++){
                scanf("%d",&v);
                g[u][v]=1;
            }
        }//建图
        int cur=hungary();
        //cout<<cur<<endl;
        printf("%d
",n-cur/2);
    }
    return 0;
}