Girls and Boys（最大独立集=节点数-最大匹配数）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1068

Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7756 Accepted Submission(s): 3559Problem Descriptionthe 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 studentsthe description of each student, in the following formatstudent_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...orstudent_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

SourceSoutheastern Europe 2000

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<vector>using namespace std;const int maxn=11111;vector<int>map[maxn];int match[maxn],n,m,k,ans;bool vis[maxn];bool dfs(int u){	for(int i=0;i<map[u].size();i++)	{		if(!vis[map[u][i]])		{			vis[map[u][i]]=true;			if(match[map[u][i]]==-1||dfs(match[map[u][i]]))			{				match[map[u][i]]=u;				return true;			}		}	}	return false;}void hangry(){	memset(match,-1,sizeof(match));	ans=0;	for(int i=0;i<n;i++)	{		memset(vis,false,sizeof(vis));		if(dfs(i))		{			ans++;		}	}}int main(){	int u,v,num,i;	while(scanf("%d",&n)==1)	{		num=n;		for(i=0;i<n;i++)		map[i].clear();		while(num--)		{		scanf("%d: (%d)",&u,&m);		for(i=1;i<=m;i++)		{			scanf("%d",&v);			map[u].push_back(v);		}	   }	   hangry();	   printf("%d\n",n-ans/2);    }	return 0;}