标签：poj 匈牙利算法 二分图最大匹配
COURSES
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 18892 |
|
Accepted: 7455 |
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 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
题目链接：http://poj.org/problem?id=1469
题目大意：一些课一些人，组成一个集体，集体中每人代表每门不同的课，每门课在集体中有一名成员，问是否能组成这样的集体
题目分析：一个人可以学多门课程并代表其中的一门课程，可是一门课程只能选一个人做代表，我们以课程和人的关系建立二分图，当这个二分图的最大匹配数等于课程数，那显然是可以的，否则不可以即必有一门课没有代表
#include <cstdio>
#include <cstring>
bool g[105][305];
int cx[105], cy[305];
bool vis[305];
int n, p;
int DFS(int x)
{
for(int y = 1; y <= n; y++)
{
if(!vis[y] && g[x][y])
{
vis[y] = true;
if(cy[y] == -1 || DFS(cy[y]))
{
cx[x] = y;
cy[y] = x;
return 1;
}
}
}
return 0;
}
int MaxMatch()
{
int res = 0;
memset(cx, -1, sizeof(cx));
memset(cy, -1, sizeof(cy));
for(int i = 1; i <= p; i++)
{
if(cx[i] == -1)
{
memset(vis, false, sizeof(vis));
res += DFS(i);
}
}
return res;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
memset(g, false, sizeof(g));
scanf("%d %d", &p, &n);
for(int i = 1; i <= p; i++)
{
int cnt;
scanf("%d", &cnt);
while(cnt --)
{
int j;
scanf("%d", &j);
g[i][j] = true;
}
}
int ans = MaxMatch();
if(ans == p)
printf("YES\n");
else
printf("NO\n");
}
}
POJ 1469 COURSES (二分图最大匹配 匈牙利算法)
标签：poj 匈牙利算法 二分图最大匹配
原文地址：http://blog.csdn.net/tc_to_top/article/details/46420537
