HDU 1068 Girls and Boys 最大独立集

问题描述：
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

题目大意：

给出一些人互相认识，求一个最大的集合，里面的人互不相识。

思路：

最大独立集 = n - 单向图最大匹配/2

AC代码：

#include<bits/stdc++.h>using namespace std;#define N 1505typedef long long ll;int n;vector<int> mapp[N];int node[N];bool vis[N];void init(){    int i;    memset(node,-1,sizeof(node));    for(i=0;i<n;i++)    {        mapp[i].clear();    }}bool dfs(int x){    int i,j,k;    for(i=0;i<mapp[x].size();i++)    {        k=mapp[x][i];        if(!vis[k])        {            vis[k]=1;            if(node[k]==-1 || dfs(node[k]))            {                node[k]=x;                return 1;            }        }    }    return 0;}int main(){    int i,j,k;    while(cin>>n)    {        init();        int a,b,c;        for(i=0;i<n;i++)        {            scanf("%d: (%d)",&a,&b);            for(j=0;j<b;j++)            {                scanf("%d",&c);                mapp[a].push_back(c);                //mapp[c].push_back(a);            }        }        int ans=0;        for(i=0;i<n;i++)        {            memset(vis,0,sizeof(vis));            if(dfs(i))                ans++;        }        cout<<n-ans/2<<endl;    }    return 0;}