POJ2524，Ubiquitous Religions

Ubiquitous Religions

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 23947 Accepted: 11792

Description

There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.

You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to 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 know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at 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 two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.

Output

For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that 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

import java.util.Arrays;import java.util.Scanner;public class poj2524 {public static int[] f;public static int n;public static void main(String[] args) {Scanner cin = new Scanner(System.in);int k = 0;n = cin.nextInt();int m = cin.nextInt();while (n != 0 && m != 0) {f = new int[n + 6];for (int i = 0; i <= n; i++) {f[i] = i;//初始化父节点为自身}for (int i = 1; i <= m; i++) {int a = cin.nextInt();int b = cin.nextInt();a = find(a);//找到祖先节点b = find(b);f[a] = b;//将a的父节点定为b，将a连接到b上}for (int i = 1; i <= n; i++) {find(i);//连接剩余可以再连接的节点}int sum = 0;Arrays.sort(f, 1, n + 1);for (int i = 1; i <= n; i++) {if (f[i] != f[i - 1])sum++;}System.out.println("Case " + (++k) + ": " + sum);n = cin.nextInt();m = cin.nextInt();}}static int find(int x) {if (f[x] != x) {//若有父节点,即x不是祖先节点//将x的父节点赋值为x的祖先节点//并将父节点到祖先节点的途中的节点全部连接到祖先节点上f[x] = find(f[x]);}return f[x];//此时的f[x]已经是x的祖先节点}}

并查集

题目大意：输入两个整数n和m，n表示有n个学生，下面有m行，每行两个数表示两个学生信仰同一个宗教，两个0是结束标记，输出学校最多有几个宗教