"Original" POJ-----2524 Ubiquitous Religions Problem Solving report

Title Address:

http://poj.org/problem?id=2524

Topic content:

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 26119		Accepted: 12859

Description

There is, many different religions in the world today that it's difficult to keep track of them all. You is interested in finding-many different religions students in your university believe in.

You know that there is n students in your university (0 < n <= 50000). It is infeasible for your to ask every student their religious beliefs. Furthermore, many students is not comfortable expressing their beliefs. One-to-avoid these problems is-ask m (0 <= m <= N (n-1)/2) pairs of students and ask them whether they believe In the same religion (e.g they may know if they both attend the same church). From this data, you may not be know what is believes in, but can get a idea of the the upper bound of how many diff Erent religions can is possibly represented on campus. You may assume the student subscribes to the most one religion.

Input

The input consists of a number of cases. Each case starts with a line specifying the integers n and M. The next m lines each consists of both integers I and j, specifying that students I and J believe in the same religion. The students is numbered 1 to N. The end of input is specified by a line in which n = m = 0.

Output

For each test case, print to a single line the case number (starting with 1) followed by the maximum number of different R Eligions the students in the university believe in.

Sample Input

10 91 21 31 41 51 61 71 81 91 1010 42 34 54 85 80 0

Sample Output

Case 1:1case 2:7

Hint

Huge input, scanf is recommended.

Source

Alberta Collegiate Programming Contest 2003.10.18 problem solving ideas: water problem. Very simple and check the set of problems, is to see people on the merger, the final statistics have several sets and then output. How do I calculate a total of several sets? As you can see, there are n individuals, so there are N collections at the time of initialization. Currently two people if need to merge, then n-1, if not need, skip directly.  The final output is left with a few sets of answers. All code:

#include <stdio.h>int Set[50001];intn,m;intFind_father (intson) {  if(Set[son] = =0) {    returnSon; }  return Set[Son] = Find_father (Set[son]);voidinit () { for(inti =0; I < n; i + +) {    Set[I] =0; }}intMainvoid){  intCount =1; intman1,man2;  while(SCANF ("%d%d", &n, &m)) {if(M = =0&& n = =0)       Break; intres =N;  for(inti =0; I < m; i + +) {scanf ("%d%d", &man1, &man2); intFAT1 =Find_father (MAN1); intFAT2 =Find_father (man2); if(FAT1! =fat2) {    Set[FAT2] =FAT1; Res--;//merge once to reduce a collection}} printf ("Case %d:%d\n", Count, res);    Init (); Count++; }  return 0;}

"Original" POJ-----2524 Ubiquitous Religions Problem Solving report