TZOJ 1321 Girls and Boys(匈牙利最大独立集)

描述

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.

输入

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 students.

输出

For each given data set, the program should write to standard output a line containing the result.

样例输入

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

样例输出

5
2

题意

N个人，求最大子集使得任何两个人都没有亲密关系

题解

先把亲密关系连图，跑匈牙利得到最大匹配，这里不知道男女，所以最大匹配得/2

答案就是N-最大匹配/2

代码

#include<bits/stdc++.h>
#define pb push_back
using namespace std;

const int N=1005;

vector<int> G[N];
int n,match[N],vis[N];
bool dfs(int u)
{
    for(int i=0;i<(int)G[u].size();i++)
    {
        int v=G[u][i];
        if(!vis[v])
        {
            vis[v]=1;
            if(match[v]==-1||dfs(match[v]))
            {
                match[v]=u;
                return true;
            }
        }
    }
    return false;
}
int hungary()
{
    int ans=0;
    memset(match,-1,sizeof match);
     for(int i=0;i<n;i++)
     {
        memset(vis,0,sizeof vis);
        ans+=dfs(i);
    }
    return ans;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)G[i].clear();
        for(int i=0,x,y,k;i<n;i++)
        {
            scanf("%d: (%d)",&x,&k);
            while(k--)scanf("%d",&y),G[x].pb(y);
        }
        printf("%d\n",n-hungary()/2);
    }
    return 0;
}