HDU 1083

        这是一道二分图最大匹配的问题，用匈牙利算法就可以解决。

        如果不懂可以参考一下：匈牙利算法。

        在写代码的过程中犯了个错误，代码中已经注释出来了，以后得小心，不要犯同样的错误。顺便说一下，题目里提到了从文件读取数据，但其实仍然是标准输入。

代码（G++）：

//链式前向星版 #include <cstdlib>#include <iostream>#define MAX 305using namespace std;struct edge{    int to;    int next;   };edge array[MAX*100];int head[MAX],cx[MAX],cy[MAX];  //cx代表课程，cy代表学生 bool path[MAX];     //记录学生集合中的点是否在交错路中            int findpath(int a){    int i;    for(i=head[a];i!=-1;i=array[i].next)    {        if(path[array[i].to]==false)        {            path[array[i].to]=true;                                    if(cy[array[i].to]==-1||findpath(cy[array[i].to]))            {                cx[a]=array[i].to;                        cy[array[i].to]=a;                                                 return 1;            }           // path[array[i].to]=false;  //这句不能有，会拖慢速度，在某些情况下会超时，但不会导致错误         }    }    return 0;}int main(int argc, char *argv[]){    int t,p,n,m,i,b,c,ans;    //c记录边数 ，m记录每门课的人数，ans记录匹配的点数 ，b记录学生编号     cin>>t;    while(t--)    {       cin>>p>>n;       memset(head,-1,sizeof(head));       memset(cx,-1,sizeof(cx));       memset(cy,-1,sizeof(cy));       c=ans=0;        for(i=1;i<=p;i++)       {          cin>>m;          while(m--)          {              cin>>b;               array[c].to=b;              array[c].next=head[i];              head[i]=c;              c++;               }       }       for(i=1;i<=p;i++)       {          if(cx[i]==-1)          {              memset(path,false,sizeof(path));                       ans+=findpath(i);          }       }       if(ans==p) cout<<"YES"<<endl;       else cout<<"NO"<<endl;    }    system("PAUSE");    return EXIT_SUCCESS;}/*邻接矩阵版 #include <cstdlib>#include <iostream>#define MAX 305using namespace std;int cx[MAX],cy[MAX],n;  //cx代表课程，cy代表学生 bool path[MAX],matrix[MAX][MAX];     //path记录学生集合中的点是否在交错路中            int findpath(int a){    int i;    for(i=1;i<=n;i++)    {        if(matrix[a][i]&&path[i]==false)        {            path[i]=true;                                    if(cy[i]==-1||findpath(cy[i]))            {                cx[a]=i;                        cy[i]=a;                                                 return 1;            }            path[i]=false;        }    }    return 0;}int main(int argc, char *argv[]){    int t,p,m,i,b,ans;    //c记录边数 ，m记录每门课的人数，ans记录匹配的点数 ，b记录学生编号     cin>>t;    while(t--)    {       cin>>p>>n;        memset(matrix,false,sizeof(matrix));             memset(cx,-1,sizeof(cx));       memset(cy,-1,sizeof(cy));       ans=0;        for(i=1;i<=p;i++)       {          cin>>m;          while(m--)          {              cin>>b;               matrix[i][b]=true;          }       }       for(i=1;i<=p;i++)       {          if(cx[i]==-1)          {              memset(path,false,sizeof(path));                       ans+=findpath(i);          }       }       if(ans==p) cout<<"YES"<<endl;       else cout<<"NO"<<endl;    }    system("PAUSE");    return EXIT_SUCCESS;}*/

题目：

Courses

                                                                   Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is to determine whether it is possible to form a committee of exactly P students that satisfies simultaneously the conditions:

. every student in the committee represents a different course (a student can represent a course if he/she visits that course)

. each course has a representative in the committee

Your program should read sets of data from a text file. The first line of the input file contains the number of the data sets. Each data set is presented in the following format:

P N
Count1 Student1 1 Student1 2 ... Student1 Count1
Count2 Student2 1 Student2 2 ... Student2 Count2
...... 
CountP StudentP 1 StudentP 2 ... StudentP CountP

The first line in each data set contains two positive integers separated by one blank: P (1 <= P <= 100) - the number of courses and N (1 <= N <= 300) - the number of students. The next P lines describe in sequence of the courses . from course 1 to course P, each line describing a course. The description of course i is a line that starts with an integer Count i (0 <= Count i <= N) representing the number of students visiting course i. Next, after a blank, you'll find the Count i students, visiting the course, each two consecutive separated by one blank. Students are numbered with the positive integers from 1 to N.

There are no blank lines between consecutive sets of data. Input data are correct.

The result of the program is on the standard output. For each input data set the program prints on a single line "YES" if it is possible to form a committee and "NO" otherwise. There should not be any leading blanks at the start of the line.

An example of program input and output:

 

Sample Input

23 33 1 2 32 1 21 13 32 1 32 1 31 1

 

Sample Output

YESNO