poj 1466 HDU 1068 Girls and Boys （最大独立集）

Girls and Boys

Description

In 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.

Input

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 (n <=500 ), for n subjects.

Output

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

Source

Southeastern Europe 2000

一群男女搞暧昧，求一个最大的集合，元素满足任意两个元素都不暧昧。

额，贴完去吃饭饭~~：

/*=============================================================================
#
#      Author: liangshu - cbam
#
#      QQ : 756029571
#
#      School : 哈尔滨理工大学
#
#      Last modified: 2015-08-27 17:19
#
#     Filename: C.cpp
#
#     Description:
#        The people who are crazy enough to think they can change the world, are the ones who do !
=============================================================================*/
#

#include<iostream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cctype>
#include<string>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;
const int INF=800;
int n;
int from[INF],tot;
bool use[INF];
vector<int>G[INF];
bool match(int x)
{
     for(int i=0;i<G[x].size();i++)
    {
         if(!use[G[x][i]])
        {
             use[G[x][i]]=1;
            if(from[G[x][i]]==-1||match(from[G[x][i]]))
            {
                 from[G[x][i]]=x;
                return 1;
            }
        }
    }
    return 0;
}

int hungary( )
{
     tot=0;
    memset(from,255,sizeof(from));
    for(int i=0;i<n;i++)
    {
         memset(use,0,sizeof(use));
        if(match(i))
           {
               ++tot;
           }
    }
    return tot;
}

int main(){
    int x, y;
    while(cin>>n){
    for(int i = 0; i < n; i++){
        scanf("%d: (%d)",&x, &y);
        for(int j = 0; j < y; j++){
            int a;
            scanf("%d",&a);
            G[x].push_back(a);
        }
    }
        cout<<n - hungary() / 2<<endl;
        for(int i = 0; i < INF; i++){
            G[i].clear();

        }
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。