HDU 1083 Courses 【二分图完备匹配】

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探讨如何使用匈牙利算法解决课程与学生匹配问题，确保每门课程都有唯一学生代表，通过二分图最大匹配实现。

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1083

Courses

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11353 Accepted Submission(s): 5326

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

3 3

3 1 2 3

2 1 2

1 1

3 3

2 1 3

1 1

Sample Output

YES

Source

Southeastern Europe 2000

题意概括：

给出 P 个课程 N 个学生，和选第 i 个课程的学生。询问是否所有课程都能被学生匹配，上每门课的学生都不同。

解题思路：

二分图最大匹配（匈牙利算法），判断最大匹配数是否等于课程数 P。

AC code:

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 using namespace std;
 9 const int MAXN = 310;
10 const int MAXP = 110;
11 
12 struct Edge
13 {
14     int v, nxt;
15 }edge[MAXP*MAXN];
16 int head[MAXN], cnt;
17 int linker[MAXN];
18 bool used[MAXN];
19 int N, P;
20 
21 void add(int from, int to)
22 {
23     edge[cnt].v = to;
24     edge[cnt].nxt = head[from];
25     head[from] = cnt++;
26 }
27 
28 bool Find(int x)
29 {
30     int v;
31     for(int i = head[x]; i != -1; i = edge[i].nxt){
32         v = edge[i].v;
33         if(used[v]) continue;
34         used[v] = true;
35         if(linker[v] == -1 || Find(linker[v])){
36             linker[v] = x;
37             return true;
38         }
39     }
40     return false;
41 }
42 
43 void init()
44 {
45     memset(head, -1, sizeof(head));
46     memset(linker, -1, sizeof(linker));
47     memset(edge, 0, sizeof(edge));
48     cnt = 0;
49 }
50 
51 int main()
52 {
53     int T_case, bnum, v;
54     scanf("%d", &T_case);
55     while(T_case--){
56         init();
57         scanf("%d%d", &P, &N);
58         for(int i = 1; i <= P; i++){
59             scanf("%d", &bnum);
60             while(bnum--){
61                 scanf("%d", &v);
62                 add(i, v);
63             }
64         }
65         int res = 0;
66         for(int i = 1; i <= P; i++){
67             memset(used, 0, sizeof(used));
68             if(Find(i)) res++;
69         }
70         if(res == P) puts("YES");
71         else puts("NO");
72     }
73     return 0;
74 }