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

Sample Output

Source

Waterloo Local 2002.01.26

 1 #include<stdio.h>
 2 #include<set>
 3 #include<string.h>
 4 using namespace std;
 5 
 6 set <int> a[110] ;
 7 bool flag [110] ;
 8 int t , p ;
 9 //set <int> :: iterator it  ;迭代器
10 
11 int main ()
12 {
13     //freopen ("a.txt" , "r" , stdin ) ;
14     memset (flag , 0 , sizeof(flag) ) ;
15     scanf ("%d%d" , &t , &p ) ;
16     int i , j ;
17     int cnt = 0 ;
18     while (~ scanf ("%d%d" , &i , &j) ) {
19         a[i].insert (j) ;
20     }
21     for (int i = 1 ; i < p ; i++) {
22         if (flag[i] == 0) {
23             flag[i] = 1 ;
24             for (int j = i + 1 ; j <= p ; j++) {
25                 if (a[i] == a[j]) {
26                     flag [j] = 1 ;
27                 }
28             }
29             cnt ++ ;
30         }
31     }
32     printf ("%d
" , cnt ) ;
33     return 0 ;
34 }

View Code

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") ;

}