ZOJ 1137二分图匹配最大独立集

Girls and Boys Time Limit: 10 Seconds      Memory Limit: 32768 KB

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.

An example is given in Figure 1.

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

 

 

 

/******************************************************
* 思路：求二分图的最大独立点集 ,
* 用匈牙利算法先求出最大匹配数 ,
* 然后用总数减去最大匹配数就是题目要求的答案
* 难度：低
* 解题要点:掌握二分图基本性质即可；
* 目测在写一下邻接表版本还有DAG的拆点建图这部分就完了
* 向网络流进军，哈哈
* author：sgx
* date:2013/09/06
*******************************************************/

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>

#define A system("pause")
using namespace std;

const int maxn=1000+5;

int map[maxn][maxn],Linker[maxn];
bool used[maxn];

int u,cnt,v;

int DFS(int u)
{
     for(int v=0;v<cnt;v++)
     {
         if(map[u][v]&&!used[v])
         {
             used[v]=true;
             if(Linker[v]==-1||DFS(Linker[v]))
             {
                 Linker[v]=u;
                return 1;
             }
         }
    }
    return 0;
}

int Hungary()
{
     int res=0;
     memset(Linker,-1,sizeof(Linker));
     for(int i=0;i<cnt;i++)
     {
         //A;
         memset(used,0,sizeof(used));
         if(DFS(i))
             res++;
     }
     return res;
}

int main()
{
     int cnt1;
     while(scanf("%d",&cnt)!=EOF)
     {
         memset(map,0,sizeof(map));
         for(int i=0;i<cnt;i++)
         {
             scanf("%d: (%d)",&u,&cnt1);
             while(cnt1--)
             {
                 scanf("%d",&v);
                 map[u][v]=1;
             }
         }
        //cout<<234<<endl;
         int res=cnt-Hungary()/2;
         printf("%d\n",res);
     }
     return 0;
}