POJ 2664 Prerequisites?

Prerequisites?

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 3342 Accepted: 2093

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 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

Waterloo local 2005.09.24

解题思路：对于选定的课，若在每个方面的必修课均被选到则输出yes,否则输出no，水题而已，直接模拟即可，我加了一个小小的优化，就是每个方面的时候，若必修课数已被修满，则后序的必修课就不必理会，奇怪的是我用动态数组就WA，用静态则AC，害得我多了几次WA，= =!,也不知道为什么。。。

#include<iostream>using namespace std;int main(){int i,j,k,select,kind,num,m,temp,number[105],major[105][105];bool pass[105],flag;while(cin>>select&&select){cin>>kind;flag=true;for(i=1;i<=select;i++)cin>>number[i];memset(pass,false,sizeof(pass));for(j=1;j<=kind;j++){cin>>num>>m;for(k=1;k<=num;k++){cin>>major[j][k];for(i=1;i<=select&&m;i++)

                                //m等于0就是必修学分全部到手，后续必修课可以不必理会

    if(number[i]==major[j][k])m--;}if(!m)pass[j]=true;}for(k=1;k<=kind;k++)if(!pass[k]){flag=false;break;}if(flag)cout<<"yes\n";elsecout<<"no\n";}return 0;}