G-圆组

题目描述

给出n个圆的圆心和半径，相交的圆算在同一组中，如圆1和圆2和圆3相交，则圆1,2,3在同一组中。求总共有几组圆。

输入

多组输入，第一行输入n，表示有n$（0<=n<=1000）$个圆，接下来n行，每行输入 圆心坐标 x，y，半径$r$ （都是int型）

输出

对每组输入输出总共圆的组数

样例输入

4
2 0 1
0 2 1
-2 0 1
0 -2 1

样例输出

4

题解

import java.util.Scanner;

public class Main {
    static int[] F = new int[1010];

    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        G[] g = new G[n];
        for (int i = 0; i < n; i++) {
            int x = cin.nextInt();
            int y = cin.nextInt();
            int r = cin.nextInt();
            g[i] = new G(x, y, r);
        }
        for (int i = 0; i < F.length; i++) {
            F[i] = i;
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (charge(g[i], g[j]) && i != j) {
                    union(i, j);
                }
            }
        }
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (F[i] == i) {
                count++;
            }
        }
        System.out.println(count);
        cin.close();
    }

    private static void union(int a, int b) {
        int fa = find(a);
        int fb = find(b);
        if (a != b) {
            F[fb] = fa;
        }
    }

    private static int find(int a) {
        return a == F[a] ? a : (F[a] = find(F[a]));
    }

    private static boolean charge(G a, G b) {
        return dis(a, b) < (a.r + b.r);
    }

    private static double dis(G a, G b) {
        double s = Math.sqrt(1.0 * (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
        return s;
    }

    private static class G {
        int x;
        int y;
        int r;

        public G(int x, int y, int r) {
            this.x = x;
            this.y = y;
            this.r = r;
        }
    }
}