题意描述：有p门课程，n个学生，每门课程可能有不止一名学生选，每个学生在一个committee代表一门课程，求学生和课程的最大匹配，若最大匹配等于p，输出YES，否则输出NO。
解题思路：入门级的二分图匹配问题，对课程—学生关系建立一个图，进行二分图的最大匹配，看最大匹配数是否等于课程数即可，使用模板即可。
错误分析：要注意输入的序号。
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:Input
2 3 3 3 1 2 3 2 1 2 1 1 3 3 2 1 3 2 1 3 1 1Output
YES NO
AC代码
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int e[310][310],book[310],match[310];
int p,n;
int dfs(int u)
{
int i;
for(i=1;i<=n;i++)
{
if(book[i]==0&&e[u][i]==1)
{
book[i]=1;
if(match[i]==-1||dfs(match[i]))
{
match[i]=u;
return 1;
}
}
}
return 0;
}
int main()
{
int t,i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&p,&n);
memset(e,0,sizeof(e));
for(i=1;i<=p;i++)
{
int k;
scanf("%d",&k);
for(j=1;j<=k;j++)
{
int m;
scanf("%d",&m);
e[i][m]=1;
}
}
memset(match,-1,sizeof(match));
int sum=0;
for(i=1;i<=p;i++)
{
memset(book,0,sizeof(book));
if(dfs(i))
sum++;
}
if(sum==p)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}