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 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
題意：給出課程數跟學生數，以及課程對應的學生。
求是否存在讓所有課程跟不同的學生對應。
#include <stdio.h>
#include <stdlib.h>
#define maxn 310
int map[maxn][maxn], hash[maxn], pre[maxn], gn, gm;
//gn左邊x的點數，gm右邊y的點數
//pre是右邊節點連接x中的節點
//hash是右邊的節點有沒有被訪問過
void set()
{
for (int i = 1; i <= gn; i++)
{
for (int j = 1; j <= gm; j++)
{
map[i][j] = 0;
}
}
for (int j = 1; j <= gm; j++)
{
pre[j] = -1;
}
}
int han(int u)
{
for (int v = 1; v <= gm; v++)
{
if (hash[v] == 0 && map[u][v])//邊存在且右節點沒被訪問過
{
hash[v] = 1;
if (pre[v] == -1 || han(pre[v]))//v還沒有前置節點或v的前置節點存在增廣路
{
pre[v] = u;//設定v的前置節點
return 1;
}
}
}
return 0;
}
int main()
{
int t, i, j, m, u, v, num;
scanf("%d", &t);
while (t--)
{
scanf("%d%d", &gn, &gm);
set();
for (i = 1; i <= gn; i++)
{
scanf("%d", &m);
while (m--)
{
scanf("%d", &v);
map[i][v] = 1;
}
}
for (i = 1, num = 0; i <= gn; i++)
{
for (j = 1; j <= gm; j++)
{
hash[j] = 0;
}
if (han(i))
{
num++;
}
}
if (num == gn)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
system("pause");
return 0;
}