贪心思想:尽可能使一个雷达能辐射到更多的岛屿
核心难点:将求雷达的辐射面积（圆）转换成求岛屿可以被辐射的海岸线范围（线）

性能瓶颈：排序
时间复杂度：O(NlogN)

JAVA代码：

import java.util.Arrays;
import java.util.Scanner;

public class Main { 

    static final int N = 1010;
    static int n, d;
    static double segs[][] = new double[N][2];

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        n = in.nextInt();
        d = in.nextInt();
        for (int i = 0; i < n; i++) {
            int x = in.nextInt(), y = in.nextInt();
            if (y > d) {
                System.out.println(-1);
                return;
            }
            double side = Math.sqrt(d * d - y * y);
            segs[i][0] = x - side;
            segs[i][1] = x + side;
        }

        Arrays.sort(segs, 0, n,(a, b)->Double.compare(a[1], b[1]));  // JDK 1.7

        int res = 0;
        double last = - 0x7fffffff;

        for (double[] seg : segs) {
            if (seg[0] > last) {
                res ++ ;
                last = seg[1];
            }
        }

        System.out.println(res);
    }

}

C++代码：

#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 1010;

int n, a[N];

struct Node
{
    double l, r;
}seg[N];

// 以x y 为最远辐射距离 确定雷达的范围：x +- sqrt(r * r - y * y)
// 求的就是这n段范围 能用几个雷达覆盖住：n - 线段交集数

int main()
{
    int n, d;
    cin >> n >> d;
    for (int i = 0; i < n; i ++ )
    {
        int x, y;
        cin >> x >> y;
        if (y > d)
        {
            cout << - 1;
            return 0;
        }
        seg[i] = {x - sqrt(d * d - y * y), x + sqrt(d * d - y * y)}; // cpp 11 语法
    }

    sort(seg, seg + n, [](Node u, Node v) {return u.r < v.r;});

    // for (int i = 0; i < n; i ++ ) cout << seg[i].l << ' ' << seg[i].r << endl;

    double last = - 0x7fffffff;
    int cnt = 0;

    for (int i = 0; i < n; i ++ )
    {
        if (seg[i].l > last)
        {
            cnt ++ ;
            last = seg[i].r;
        }
    }

    cout << cnt << endl;

    return 0;

}

萌新，请多多支持！QAQ