Summer Earnings

在一个平面内给出nn个点的坐标，任选其中三个为圆心作半径相同的圆，要求这三个圆不能相交但可以相切，求能画出的圆中的最大半径。

分析

待补充

AC

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.BitSet;

public class Main {
	public static class Edge implements Comparable {
		int len;
		int a, b;//端点
		public Edge(int a, int b) {
			this.a = a;
			this.b = b;
			this.len = (x[a] - x[b]) * (x[a] - x[b]) + (y[a] - y[b]) * (y[a] - y[b]);
		}
		@Override
		public int compareTo(Object o) {
			return ((Edge)o).len - this.len;
		}
	}
	static int[] x = new int[3010];
	static int[] y = new int[3010];
	static Edge[] edges = new Edge[4500000];
	public static void main(String[] args) throws IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(bf.readLine());
		
		for(int i = 0; i < n; ++i) {
			String[] strs = bf.readLine().split(" ");
			x[i] = Integer.parseInt(strs[0]);
			y[i] = Integer.parseInt(strs[1]);
		}
		int nums = 0;
		for(int i = 0; i < n; ++i) {
			for(int j = i + 1; j < n; ++j) {
				edges[nums++] = new Edge(i, j);
			}
		}
		Arrays.sort(edges, 0, nums);
		BitSet[] bitSets = new BitSet[n];
		for(int i = 0; i < n; ++i) {
			bitSets[i] = new BitSet(3010);
		}
		for(int i = 0; i < nums; ++i) {
			BitSet tmp = (BitSet) bitSets[edges[i].a].clone();
			tmp.and(bitSets[edges[i].b]);
			if(tmp.cardinality() != 0) {
				System.out.printf("%.20f\n", Math.sqrt(edges[i].len) / 2.0);
				break;
			} else {
				bitSets[edges[i].a].set(edges[i].b, true);
				bitSets[edges[i].b].set(edges[i].a, true);
			}
		}
		
	}
}

链接：https://www.luogu.com.cn/problem/CF333E
来源：洛谷