hdu 1144 Prerequisites

题目地址：

http://acm.hdu.edu.cn/showproblem.php?pid=1144

题目描述：

Prerequisites?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 900    Accepted Submission(s): 561

Problem Description

Freddie the frosh has chosen to take k courses. To meet the degree requirements, he must take courses from each of several categories. Can you assure Freddie that he will graduate, based on his course selection? 

 

Input

Input consists of several test cases. For each case, the first line of input contains 1 ≤ k ≤ 100, the number of courses Freddie has chosen, and 0 ≤ m ≤ 100, the number of categories. One or more lines follow containing k 4-digit integers follow; each is the number of a course selected by Freddie. Each category is represented by a line containing 1 ≤ c ≤ 100, the number of courses in the category, 0 ≤ r ≤ c, the minimum number of courses from the category that must be taken, and the c course numbers in the category. Each course number is a 4-digit integer. The same course may fulfil several category requirements. Freddie's selections, and the course numbers in any particular category, are distinct. A line containing 0 follows the last test case.

 

Output

For each test case, output a line containing "yes" if Freddie's course selection meets the degree requirements; otherwise output "no." 

 

Sample Input

   
   
   
   

    
    
    
    3 2
0123 9876 2222
2 1 8888 2222
3 2 9876 2222 7654 
3 2
0123 9876 2222
2 2 8888 2222
3 2 7654 9876 2222
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    yes
no
   
   
   
   

题意：

选课，要选k门课，每门课给出课程号，有m类课程，每类课程有c个课程，在该类课程至少要选r门课程才能达到学分，问给出相应课程，课程类，课程号是否能达到学分。

题解：

每一类只要满足r门课程的数量即可，即每个样例的第3～m行 每行都要满足r门课程的选择就可通过，只要有一行没有通过的就是不能通过。而满足这个r门课程根据课程号来匹配，是否在3～m行选的这些课程号涉及到了第2行的课程，若涉及到了说明满足一门，所以passcnt++（见代码）

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
using namespace std;
int course[100]={0};
int k=0,m=0,c=0,r=0;
int cates[100]={0};
int main()
{
    while(scanf("%d",&k)!=EOF&&k>0)
    {
        int flag=1;//initialize the case is available
        scanf("%d",&m);
        for(int i=0;i<=k-1;i++)
        {
            scanf("%04d",&course[i]);
        }
        for(int i=0;i<=m-1;i++)
        {
            scanf("%d%d",&c,&r);
            int passcnt=0;
            for(int j=0;j<=c-1;j++)
            {
                int degree=0;
                scanf("%d",°ree);
                for(int x=0;x<=k-1;x++)
                {
                    if(course[x]==degree)
                    {
                        passcnt++;
                    }
                }
            }
            if(passcnt<r)
            {
                flag=0;
            }
        }
        if(!flag)
        {
            printf("no\n");
        }
        else
        {
            printf("yes\n");
        }
    }
    return(0);
}