785
【二分匹配】HDU1083-Courses
Courses
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3645 Accepted Submission(s): 1745
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
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
p門課，每門課有若幹名學生喜歡，一共有N個學生，每個學生可以喜歡多門課，寫出每門課以及喜歡這門課的學生，
求是否能為所有的課安排班長（一個人隻能任命為一門課的班長），如果能輸出“YES”，不能輸出“NO”
思路：
將學生看成A集合，課程看成B集合，即求A集合中給的元素（數量可多於B集合）是否能夠
完全匹配上B集合的所有元素（每個A隻能和B中某些元素匹配）
明顯的二分匹配，用匈牙利算法
AC代碼：
#include<stdio.h>
#include<string.h>
int course[510],flag[510],line[510][510];//被HDU坑了，數組顯然比100和300要大才能AC
int n,m;
int Find(int x)
{
int i;
for(i=1;i<=m;i++)//遍曆所有被選課
{
if(line[x][i]==1&&flag[i]==0)
{//如果 x喜歡i課且在這一個遞歸選取階段沒有被選取（哪怕是暫時選取，新的遞歸可能會換）
flag[i]=1;//標記被選取
if(course[i]==0||Find(course[i]))//如果被選課沒有班長或它的班長可以調換（它的班長可以選擇其它被選課）
{
course[i]=x;//將i課的班長定為 x
return 1;
}
}
}
return 0;
}
int main()
{
int i,j,T,x,y,k,sum;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&m,&n);//m是課程數，n是學生數目，別弄反了！！！
memset(line,0,sizeof(line));
memset(course,0,sizeof(course));
for(i=1;i<=m;i++)
{
scanf("%d",&x);
while(x--)
{
scanf("%d",&y);
line[y][i]=1;//表示 y學生喜歡 i課
}
}
sum=0;//記錄能選的班長數
for(i=1;i<=n;i++)
{
memset(flag,0,sizeof(flag));//每次都要清 0
if(Find(i)) sum++;//找到一對就記錄
}
if(sum==m)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
最後更新：2017-04-03 05:39:54