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
2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1
Sample Output
YES
NO
問題不難！關鍵在於算法！這種題型屬於二分圖的匹配問題！用匈牙利算法來解答最好不過！關於匈牙利算法的描述，請看這裡，講的不能再好了，仔細看，會很有收獲的！
我花了幾個小時，終於把這個算法弄懂了！很強大的算法！沒想到離散數學竟然也這麼重要！呵！
貼代碼吧！
[cpp]
//現在弄到一個判斷二分匹配的算法，仔細研究研究，很爽的！
#include<iostream>
using namespace std;
const int Max=500;
int g[Max][Max];
int linker[Max];
bool used[Max];
bool dfs(int u)
{
int v;
for(v=1;v<=g[0][1];v++)
{
if(g[u][v] && !used[v])
{
used[v]=true;
if(linker[v]==-1 || dfs(linker[v]))
{
linker[v]=u;
return true;
}
}
}
return false;
}
int hungary()
{
int res=0;
int u;
memset(linker,-1,sizeof(linker));
for(u=1;u<=g[0][0];u++)
{
memset(used,0,sizeof(used));
if(dfs(u)) res++;
}
return res;
}
int main()
{
int cases;
cin>>cases;
while(cases--)
{
int num,m;
memset(g,0,sizeof(g));
cin>>g[0][0]>>g[0][1];
for(int i=1;i<=g[0][0];i++)
{
cin>>num;
for(int j=0;j<num;j++)
{
cin>>m;
g[i][m]=1;
}
}
if(hungary()!=g[0][0])
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
}
return 0;
}
