Poj 1106 Transmitters

最新推荐文章于 2025-09-06 09:33:29 发布

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最新推荐文章于 2025-09-06 09:33:29 发布

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文章标签： java

原文链接：http://www.cnblogs.com/zsyacm666666/p/6797289.html

本文介绍了一种解决PoJ1106题目的算法——如何找到一个可旋转的半圆能覆盖的最大点数。通过枚举每个点作为必选点，并利用叉积来判断其它点是否能在同一半圆内，从而求得最优解。

Poj 1106 Transmitters

传送门

给出一个半圆，可以任意旋转，问这个半圆能够覆盖的最多点数。
我们枚举每一个点作为必然覆盖点，那么使用叉积看极角关系即可判断其余的点是否能够与其存在一个半圆内



import java.io.*;
import java.util.*;

public class Main {
    static class Point implements Comparable<Point> {
        double x, y;

        @Override
        public int compareTo(Point a) {
            return (int) cross(a);
        }

        public double cross(Point a) {
            return a.x * y - x * a.y;
        }
    }

    static double cross(Point st, Point a, Point b) {
        return (a.x - st.x) * (b.y - st.y) - (b.x - st.x) * (a.y - st.y);
    }

    static double dist(Point a1, Point a2) {
        return (a1.x - a2.x) * (a1.x - a2.x) + (a1.y - a2.y) * (a1.y - a2.y);
    }

    static final int N = 10005;
    static final int inf = 0x3f3f3f3f;
    static final double eps = 1e-9;
    static Point a[] = new Point[N];

    public static void main(String[] args) {
        Scanner cin = new Scanner(new InputStreamReader(System.in));
        while (cin.hasNext()) {
            Point st = new Point(), tmp = new Point();
            st.x = cin.nextDouble();
            st.y = cin.nextDouble();
            double r = cin.nextDouble();
            double d;
            int n = cin.nextInt();
            int cnt = 0;
            for (int i = 0; i < n; i++) {
                tmp.x = cin.nextDouble();
                tmp.y = cin.nextDouble();
                d = dist(st, tmp);
                if (d < r * r || Math.abs(d - r * r) < eps) {
                    cnt++;
                    if (a[cnt] == null)
                        a[cnt] = new Point();
                    a[cnt].x = tmp.x;
                    a[cnt].y = tmp.y;
                }
            }
            int ans = 0;
            for (int i = 1; i <= cnt; i++) {
                int count = 1;
                for (int j = 1; j <= cnt; j++) {
                    if (i != j && cross(st, a[i], a[j]) <= 0)
                        count++;
                }
                if (count > ans)
                    ans = count;
            }
            System.out.println(ans);
        }
        cin.close();
    }
}

转载于:https://www.cnblogs.com/zsyacm666666/p/6797289.html