POJ2524，Ubiquitous Religions

Description

Input

Output

Sample Input

10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0

Sample Output

Case 1: 1
Case 2: 7

Hint

Source

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

POJ2524，Ubiquitous Religions,布布扣,bubuko.com

POJ2524，Ubiquitous Religions

原文：http://blog.csdn.net/ieayoio/article/details/38259031