POJ 1469 COURSES

Posted 2020-08-05

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21382		Accepted: 8408

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 

Input

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

P N 
Count1 Student_{1 1} Student_{1 2} ... Student_{1 Count1} 
Count2 Student_{2 1} Student_{2 2} ... Student_{2 Count2} 
... 
CountP Student_{P 1} Student_{P 2} ... Student_{P 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抣l 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. 

Output

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.

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

题目大意

N个学生，p们课，每一门课程可以有多个学生选修，给出每门课程学生的选修情况，问：对于每门课程可不可以找到一个课代表（ 前提是该学生要选修该门课程，每个课程都有一个不同的课代表）

输入：

输入有多组数据：

每组数据的第一行包含两个由空格隔开的两个正整数：P（1 < = P < = 100）课程数量和N（1 < = 300）学生人数。

以下P行描述从课程1到课程P：第一个数是选修人数Ci，后面Ci个学生编号。

输出：YES 或 NO

 

总之就是二分图的最大匹配问题

匈牙利算法走起（hangarian）

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int maxn=305;
 8 const int maxm=60005;
 9 int Head[105],Next[maxm],Adj[maxm];
10 int match[maxn];
11 bool use[maxn];
12 int c,N,P,cnt;
13 
14 void AddEdge(int u,int v)
15 {
16     c++;
17     Adj[c]=v;
18     Next[c]=Head[u];
19     Head[u]=c;
20 }
21 
22 bool dfs(int x)
23 {
24     for(int i=Head[x];i;i=Next[i])
25     {
26         int v=Adj[i]-100;
27         if(use[v]) continue;
28         use[v]=true;
29         if(match[v]==0||dfs(match[v]))
30         {
31             match[v]=x;
32             return true;
33         }
34     }
35     return false;
36 }
37 
38 void hangarian()
39 {
40     memset(match,0,sizeof(match));
41     for(int i=1;i<=P;i++)
42     {
43         memset(use,false,sizeof(use));
44         if(dfs(i)) cnt++;
45     }
46 }
47 
48 int main()
49 {
50     int t;
51     cin>>t;
52     while(t--)
53     {
54         cin>>P>>N;
55         int m,v;
56         cnt=c=0;
57         memset(Head,0,sizeof(Head));
58         memset(Next,0,sizeof(Next));
59         memset(Adj,0,sizeof(Adj));
60         for(int i=1;i<=P;i++)
61         {
62             scanf("%d",&m);
63             while(m--)
64             {
65                 scanf("%d",&v);
66                 AddEdge(i,v+100);
67             }
68         }
69         hangarian();
70         if(cnt==P) printf("YES");
71         else printf("NO");
72         cout<<endl;
73     }
74     return 0;
75 }

View Code