Poj(1469),二分图最大匹配
题目链接:http://poj.org/problem?id=1469
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 21229 | Accepted: 8355 |
Description
- 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
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
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
给你p门课程和n个学生,一个学生可以选0门,1门,或者多门课程,现在要求一个由p个学生组成的集合,满足下列2个条件:
1.每个学生选择一个不同的课程
2.每个课程都有不同的代表
如果满足,就输出YES
分析:
做最大匹配,要是匹配数是P就是YES.
#include <stdio.h>
#include <string.h> int p,n;
bool maps[][];
bool use[];
int match[]; bool DFS(int u)
{
for(int i=; i<=n; i++)
{
if(!use[i]&&maps[u][i])
{
use[i] = true;
if(match[i]==-||DFS(match[i]))
{
match[i] = u;
return true;
}
}
}
return false;
} int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(match,-,sizeof(match));
memset(maps,false,sizeof(maps));
scanf("%d%d",&p,&n);
int num = ;
for(int i=; i<=p; i++)
{
int stu;
scanf("%d",&stu);
for(int j=; j<=stu; j++)
{
int v;
scanf("%d",&v);
maps[i][v] = true;
}
} for(int i=; i<=p; i++)
{ memset(use,false,sizeof(use));
if(DFS(i))
num++;
}
if(num==p)
printf("YES\n");
else printf("NO\n"); } return ;
}
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