HDU 1083 - Courses - [匈牙利算法模板题]

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题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=1083

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
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
 
题意：
班上有N个同学，总共有P门课，给出上第Pi门课的所有学生的人数和编号；
现组建班委会，要求班委会里每门课都有一个人当课代表，并且一个人不能兼任课代表，求是否能组建起满足要求的班委会。
 
题解：
显然，将P门课作为U集，N个学生作为V集，按照上课情况得到一个二分图，如果最大匹配数正好是P，就能组建起班委会，否则就组建不了。
 
AC代码：
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<vector>
 4 #define MAX 410
 5 using namespace std;
 6 int p,n;
 7 struct Edge{
 8     int u,v;
 9 };
10 vector<Edge> E;
11 vector<int> G[MAX];
12 int matching[MAX];
13 int vis[MAX];
14 void init(int l,int r)
15 {
16     E.clear();
17     for(int i=l;i<=r;i++) G[i].clear();
18 }
19 void add_edge(int u,int v)
20 {
21     E.push_back((Edge){u,v});
22     E.push_back((Edge){v,u});
23     int _size=E.size();
24     G[u].push_back(_size-2);
25     G[v].push_back(_size-1);
26 }
27 bool dfs(int u)
28 {
29     for(int i=0,_size=G[u].size();i<_size;i++)
30     {
31         int v=E[G[u][i]].v;
32         if (!vis[v])
33         {
34             vis[v]=1;
35             if(!matching[v] || dfs(matching[v]))
36             {
37                 matching[v]=u;
38                 matching[u]=v;
39                 return 1;
40             }
41         }
42     }
43     return false;
44 }
45 int hungarian()
46 {
47     int ret=0;
48     memset(matching,0,sizeof(matching));
49     for(int i=1;i<=p;i++)
50     {
51         if(!matching[i])
52         {
53             memset(vis,0,sizeof(vis));
54             if(dfs(i)) ret++;
55         }
56     }
57     return ret;
58 }
59 int main()
60 {
61     int t;
62     scanf("%d",&t);
63     while(t--)
64     {
65         scanf("%d%d",&p,&n);
66         init(1,p+n);
67         for(int i=1,cnt,stu;i<=p;i++)
68         {
69             scanf("%d",&cnt);
70             for(int j=1;j<=cnt;j++)
71             {
72                 scanf("%d",&stu);
73                 add_edge(i,stu+p);
74             }
75         }
76         if(hungarian()==p) printf("YES\n");
77         else printf("NO\n");
78     }
79 }

 

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