poj.2419.Forests (枚举 + set用法）

Forests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5782		Accepted: 2218

Description

If a tree falls in the forest, and there's nobody there to hear, does it make a sound? This classic conundrum was coined by George Berkeley (1685-1753), the Bishop and influential Irish philosopher whose primary philosophical achievement is the advancement of what has come to be called subjective idealism. He wrote a number of works, of which the most widely-read are Treatise Concerning the Principles of Human Knowledge (1710) and Three Dialogues between Hylas and Philonous (1713) (Philonous, the "lover of the mind," representing Berkeley himself).

Input

A forest contains T trees numbered from 1 to T and P people numbered from 1 to P. Standard input consists of a line containing P and T followed by several lines, containing a pair of integers i and j, indicating that person i has heard tree j fall.

Output

People may have different opinions as to which trees, according to Berkeley, have made a sound. Output how many different opinions are represented in the input? Two people hold the same opinion only if they hear exactly the same set of trees. You may assume that P < 100 and T < 100.

Sample Input

3 4
1 2
3 3
1 3
2 2
3 2
2 4

Sample Output

2

Source

Waterloo Local 2002.01.26

 #include<stdio.h>
 #include<set>
 #include<string.h>
 using namespace std;

 set <int> a[] ;
 bool flag [] ;
 int t , p ;
 //set <int> :: iterator it  ;迭代器

 int main ()
 {
     //freopen ("a.txt" , "r" , stdin ) ;
     memset (flag ,  , sizeof(flag) ) ;
     scanf ("%d%d" , &t , &p ) ;
     int i , j ;
     int cnt =  ;
     while (~ scanf ("%d%d" , &i , &j) ) {
         a[i].insert (j) ;
     }
     for (int i =  ; i < p ; i++) {
         if (flag[i] == ) {
             flag[i] =  ;
             for (int j = i +  ; j <= p ; j++) {
                 if (a[i] == a[j]) {
                     flag [j] =  ;
                 }
             }
             cnt ++ ;
         }
     }
     printf ("%d\n" , cnt ) ;
     return  ;
 }

从师兄们那学来的orz ，首先知道了如何把set中的元素输出：

set <int> ::iterator it ;

for (it = a.begin () ; it <= a.end () ; i++ )

　　printf ("%d " , *it ) ;

----------接下来是如何判断两个set集合相等----------

简单粗暴orz：

set <int> a , b ;

if (a == b) {

　　puts ("Yes") ;

}

else {
　　puts ("No") ;

}