POJ - 1469 COURSES [二分图匈牙利算法]

COURSES

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

Source

Southeastern Europe 2000

题意:

P门课N个学生

列出每门课有几个学生上

现在每门课要有课代表，问你可不可以为每门课找到课代表。并且每个学生只能当一门课程的课代表。

做法:

裸*匈牙利算法

 1 #include<iostream>
 2 using namespace std;
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<vector>
 6 const int maxn = 1000;
 7 vector <int> maps[maxn];
 8 int n,m;
 9 int ok[maxn];//ok[i]代表第i个节点可否配对
10 int matched[maxn];//matched[i]代表第i个节点的配对对象
11 bool match(int x){
12     for(int i=0;i<maps[x].size();i++){
13         int k = maps[x][i];
14         if(ok[k]==0){
15             ok[k]=1;
16             if(matched[k]==0||match(matched[k])==true){
17                 matched[k]=x;
18                 return true;
19             }
20         }
21     }
22     return false;
23 }
24 int main(){
25     int t;
26     scanf("%d",&t);
27     while(t--&&scanf("%d%d",&n,&m)!=EOF){
28         memset(matched,0,sizeof(matched));
29         for(int i=0;i<=n;i++)
30             maps[i].clear();
31         for(int i=1;i<=n;i++){
32             int k;
33             scanf("%d",&k);
34             for(int j=0;j<k;j++){
35                 int p;
36                 scanf("%d",&p);
37                 maps[i].push_back(p);
38             }
39         }
40         int ans = 0;
41         for(int i=1;i<=n;i++){
42             memset(ok,0,sizeof(ok));
43             if(match(i)==true)
44                 ans+=1;
45         }
46         if(ans==n){
47             printf("YES\n");
48         }
49         else
50             printf("NO\n");
51     }
52     return 0;
53 }

原文地址：https://www.cnblogs.com/xfww/p/8531127.html