HDU 1068 Girls and Boys（二分匹配--匈牙利算法）

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
典型的二分匹配题目。让你求出不能组成对的最少人数，言外之意就是让你求出最大的匹配数，然后再用总人数减去匹配数的二分之一就好了。

下面是AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,m;int e[1005][1005];int book[1005];int match[1005];int dfs(int u){    for(int i=0;i<n;i++)    {        if(book[i]==0&&e[u][i]==1)        {            book[i]=1;            if(match[i]==-1||dfs(match[i]))            {                match[i]=u;                return 1;            }        }    }    return 0;}int main(){    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {            match[i]=-1;        }        memset(e,0,sizeof(e));        int t1,t2,u;        for(int i=0;i<n;i++)        {            scanf("%d: (%d)",&t1,&t2);            for(int j=0;j<t2;j++)            {                scanf("%d",&u);                e[t1][u]=1;            }        }        int sum=0;        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                book[j]=0;            }            if(dfs(i)==1)            {                sum++;            }        }        printf("%d\n",n-sum/2);    }    return 0;}