HDU1068Girls and Boys

题目：

Girls and Boys

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13942    Accepted Submission(s): 6561

 

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.

 

 

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：

有N名学生，每个学生有一组可能的和其余学生的恋爱关系名单，问最多可能组成的恋爱关系数目：

分析：

二分图的最大匹配问题

由于爱慕关系是相互的，所以匹配数会多2倍，最后结果/2

代码:

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 1000;
vector<int> G[maxn];
int vis[maxn];
int match[maxn];

void init(){
    memset(match,-1,sizeof(match));
    for(int i=0;i<maxn;i++) G[i].clear();
}

bool findpath(int k){
    for(int i=0;i<G[k].size();i++){
        int v = G[k][i];
        if(!vis[v]){
            vis[v] = 1;
            if(match[v]==-1 || findpath(match[v])){
                match[v] = k;
                return true;
            }
        }
    }
    return false;
}
int Search(int m){
    int cnt = 0;
    for(int i=0;i<m;i++){
        memset(vis,0,sizeof(vis));
        if(findpath(i)) cnt++;
    }
    return cnt;
}

int main(){
    int N;
    while(scanf("%d",&N)!=EOF){
        init();
        int T = N;
        while(T--){
            int a,b;
            scanf("%d: (%d) ",&a,&b);
            for(int i=0;i<b;i++){
                int v;
                scanf("%d",&v);
                G[a].push_back(v);
            }
        }
        int res = Search(N);
        printf("%d\n",N-res/2);
    }
    return 0;
}