ZOJ - 2548 Prerequisites?

Prerequisites?

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

Submit Status

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, and0 ≤ 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 20123 9876 22222 1 8888 22223 2 9876 2222 7654 3 20123 9876 22222 2 8888 22223 2 7654 9876 22220

Sample Output

yesno

Source

University of Waterloo Local Contest 2005.09.24

 

 

 

分析：

水题。

ac代码：

#include <iostream>
#include<cstdio>
using namespace std;
int a[101],b[101];
int main()
{
   int k,m,c,r,i,j,t;
   int count,c1;
   while(scanf("%d",&k)&&k)//
   {
       scanf("%d",&m);
       for(i=0;i<k;i++)
       scanf("%d",&a[i]);
       c1=0;
       for(i=0;i<m;i++)
       {
          scanf("%d%d",&c,&r);
          for(j=0;j<c;j++)
          scanf("%d",&b[j]);
          count=0;
          for(j=0;j<k;j++)
          {
              for(t=0;t<c;t++)
              {
                  if(a[j]==b[t])
                  count++;
              }
          }
          if(count>=r)//注意不是count==r
          {
              c1++;
          }
       }
       if(c1==m)
       {
           printf("yes\n");
       }
       else
       {
           printf("no\n");
       }
   }
    return 0;
}