标签：poj 水
COURSES
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 18993 |
|
Accepted: 7486 |
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
简单的二分匹配，将学生看成要匹配的集合X，将课程看做要匹配的集合Y，当最大匹配 == 课程数的时候，就输出YES，否则就输出NO。
代码如下：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 500;
int p, n;
int s[N][N], c[N], used[N];
bool find(int x)
{
for(int j = 1; j <= p; j++)
{
if(s[x][j] == 1 && used[j] == 0)
{
used[j] = 1;
if(c[j] == 0 || find(c[j]))
{
c[j] = x;
return true;
}
}
}
return false;
}
int main()
{
int t = 0;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &p, &n);
int x, st;
memset(s, 0, sizeof(s));
memset(c, 0, sizeof(c));
for(int i = 1; i <= p; i++)
{
scanf("%d", &x);
while(x--)
{
scanf("%d", &st);
s[st][i] = 1;
}
}
int all = 0;
for(int i = 1; i <= n; i++)
{
memset(used, 0, sizeof(used));
if(find(i))
all+= 1;
}
if(all == p) printf("YES\n");
else printf("NO\n");
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
POJ 1469（裸二分匹配）
标签：poj 水
原文地址：http://blog.csdn.net/doris1104/article/details/46728971
