前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >挑战程序竞赛系列(86):3.6极限情况(3)

挑战程序竞赛系列(86):3.6极限情况(3)

作者头像
用户1147447
发布2018-01-02 10:34:24
6740
发布2018-01-02 10:34:24
举报
文章被收录于专栏:机器学习入门

挑战程序竞赛系列(86):3.6极限情况(3)

传送门:AOJ 2201: Immortal Jewels

翻译参考至hankcs: http://www.hankcs.com/program/algorithm/aoj-2201-immortal-jewels.html

题意:

求婚:有个贵族向一个贫穷的公主求婚,公主提出条件,需要一种“永生宝石”做嫁妆。这种宝石极其稀有,而且极易损毁,所以开采时需要特别小心。如图:

矿工需要使用一种特殊的金属棒开采,宝石呈圆形,矿床是二维平面,每颗宝石坐标为x,y,半径为r,能够吸附在距离m内的金属棒上。一旦金属棒穿过某个宝石,该宝石就被破坏了。

金属棒非常昂贵,只有一根,作为贵族雇佣的程序员,请你帮他算出能开采到的最大宝石数? 数据格式如下: N x1 y1 r1 m1 x2 y2 r2 m2 … xN yN rN mN 由多个用例组成,输出答案并换行。

直接参考hankcs:

对每个宝石圆,都添加半径扩大m_i的磁力范围圆,将两个圆同时纳入考虑,移动直线的过程中,如果能取到的宝石个数发生变化,那么直线肯定与所有圆中的两个相切,这就是极限情况。所以枚举两个圆的组合确定一条直线,取最大值就行了。

此题的麻烦之处在于求解两个圆的外接切线和内接切线,具体直接参考代码吧。

代码如下:

代码语言:javascript
复制
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201709/A2201.txt";

    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    static final double PI  = Math.acos(-1);
    static final double EPS = 1E-12;

    class Point{

        double x;
        double y;

        Point(double x, double y){
            this.x = x;
            this.y = y;
        }

        Point add(Point a) {
            return new Point(x + a.x, y + a.y);
        }

        Point sub(Point a) {
            return new Point(x - a.x, y - a.y);
        }

        Point rot(double rad) {
            double a = Math.cos(rad);
            double b = Math.sin(rad);
            return new Point(x * a - b * y, x * b + a * y);
        }

        double abs() {
            return Math.sqrt(x * x + y * y);
        }
    }

    double dist(Point a, Point b) {
        double dx = a.x - b.x;
        double dy = a.y - b.y;
        return Math.sqrt(dx * dx + dy * dy);
    }

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

    class Line{

        Point a;
        Point b;

        Line(Point a, Point b){
            this.a = a;
            this.b = b;
        }

        double distance(Point p) {
            double d = dist(a, b);
            return Math.abs(dot(p.sub(a), b.sub(a)) / d);
        }
    }

    class Circle{

        Point o;
        double r;

        Circle(Point o, double r){
            this.o = o;
            this.r = r;
        }

        // 通过点p 的两条切线
        List<Point> tangent(Point p){
            double L = dist(o, p);
            double M = Math.sqrt(L * L - r * r);
            double theta = Math.asin(r / L);
            Point v = o.sub(p);
            v.x /= L;
            v.y /= L;
            List<Point> ans = new ArrayList<>();
            Point t = v.rot(theta);
            t.x *= M;
            t.y *= M;
            ans.add(p.add(t));
            t = v.rot(-theta);
            t.x *= M;
            t.y *= M;
            ans.add(p.add(t));
            return ans;
        }

        // 两个半径相等圆的两条平行外切线
        List<Line> outer_tangent_parallel(Circle c){
            Point d = o.sub(c.o);
            Point v = new Point(-r / d.abs() * d.y, r / d.abs() * d.x);
            List<Line> lines = new ArrayList<>();
            lines.add(new Line(o.add(v), c.o.add(v)));
            lines.add(new Line(o.sub(v), c.o.sub(v)));
            return lines;
        }

        // 两圆外切线
        List<Line> outer_tangent(Circle c){
            if (cmp(r, c.r) == 0) {
                return outer_tangent_parallel(c);
            }
            if (cmp(r, c.r) == 1) {
                return c.outer_tangent(this);
            }
            Point d = o.sub(c.o);
            double fact = c.r / r - 1;
            Point base = c.o.add(d).add(new Point(d.x / fact, d.y / fact));
            List<Point> ps = tangent(base);
            List<Line> ans = new ArrayList<>();
            ans.add(new Line(base, ps.get(0)));
            ans.add(new Line(base, ps.get(1)));
            return ans;
        }

        // 两圆内切线
        List<Line> inner_tangent(Circle c){
            if (cmp(r, c.r) == 1) {
                return c.inner_tangent(this);
            }
            Point d = c.o.sub(o);
            double fact = c.r / r + 1;
            Point base = o.add(new Point(d.x / fact, d.y / fact));
            List<Point> ps = tangent(base);
            List<Line> ans = new ArrayList<>();
            ans.add(new Line(base, ps.get(0)));
            ans.add(new Line(base, ps.get(1)));
            return ans;
        }

        // 两个圆的交点
        Line intersection(Circle c) {
            double d = dist(o, c.o);
            double cos = Math.cos((d * d + r * r - c.r * c.r) / (2 * d * r));
            Point e = c.o.sub(o);
            e.x /= d;
            e.y /= d;
            Point t1 = e.rot(cos);
            t1.x *= r;
            t1.y *= r;
            Point t2 = e.rot(-cos);
            t2.x *= r;
            t2.y *= r;
            return new Line(o.add(t1), o.add(t2));
        }

        // 是否相离
        boolean independent(Circle c) {
            return cmp(dist(o, c.o), r + c.r) > 0;
        }

        // 是否包含圆c
        boolean contains(Circle c) {
            return cmp(dist(o, c.o) + c.r, r) < 0;
        }

        boolean intersects(Circle c) {
            return !contains(c) && !c.contains(this) && !independent(c);
        }
    }

    class Pair{

        Circle fir;
        Circle sec;

        Pair(Circle fir, Circle sec){
            this.fir = fir;
            this.sec = sec;
        }
    }

    int cmp(double a, double b) {
        double diff = a - b;
        if (Math.abs(diff) < EPS) return 0;
        else if (diff < 0) return -1;
        else return 1;
    }

    List<Pair> jewels;
    List<Line> lines;
    int N;

    void constructLine(Circle c1, Circle c2, List<Line> lines) {
        // 两圆相交 && 两圆相离
        if (c1.independent(c2)) { // 相离
            List<Line> outer = c1.outer_tangent(c2);
            lines.add(outer.get(0));
            lines.add(outer.get(1));
            List<Line> inner = c1.inner_tangent(c2);
            lines.add(inner.get(0));
            lines.add(inner.get(1));
        }

        if (c1.intersects(c2)) {  // 相交
            List<Line> outer = c1.outer_tangent(c2);
            lines.add(outer.get(0));
            lines.add(outer.get(1));
            Line inter = c1.intersection(c2);
            lines.add(inter);
        }
    }

    int count(List<Pair> jewels, Line line) {
        int cnt = 0;
        for (Pair j : jewels) {
            if (cmp(j.fir.r, line.distance(j.fir.o)) <= 0 && cmp(j.sec.r, line.distance(j.sec.o)) >= 0) {
                cnt ++;
            }
        }
        return cnt;
    }

    void read() {
        while (true) {
            N = ni();
            if (N == 0) break;

            jewels = new ArrayList<>();
            lines  = new ArrayList<>();

            for (int i = 0; i < N; ++i) {
                double x, y, r, m;
                x = nd();
                y = nd();
                r = nd();
                m = nd();
                Pair jewel = new Pair(new Circle(new Point(x, y), r), new Circle(new Point(x, y), r + m));
                for (Pair p : jewels) {
                    constructLine(jewel.fir, p.fir, lines);
                    constructLine(jewel.fir, p.sec, lines);
                    constructLine(jewel.sec, p.fir, lines);
                    constructLine(jewel.sec, p.sec, lines);
                }
                jewels.add(jewel);
            }

            int ans = 1;
            for (Line line : lines) {
                ans = Math.max(ans, count(jewels, line));
            }
            out.println(ans);
        }
    }

    FastScanner in;
    PrintWriter out;

    void run() throws IOException {
        boolean oj;
        try {
            oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
        } catch (Exception e) {
            oj = System.getProperty("ONLINE_JUDGE") != null;
        }

        InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
        in = new FastScanner(is);
        out = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
        read();
        out.flush();
        if (!oj){
            System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
        }
    }

    public boolean more(){
        return in.hasNext();
    }

    public int ni(){
        return in.nextInt();
    }

    public long nl(){
        return in.nextLong();
    }

    public double nd(){
        return in.nextDouble();
    }

    public String ns(){
        return in.nextString();
    }

    public char nc(){
        return in.nextChar();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        boolean hasNext;

        public FastScanner(InputStream is) throws IOException {
            br = new BufferedReader(new InputStreamReader(is));
            hasNext = true;
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    hasNext = false;
                    return "##";
                }
            }
            return st.nextToken();
        }

        String next = null;
        public boolean hasNext(){
            next = nextToken();
            return hasNext;
        }

        public int nextInt() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Integer.parseInt(more);
        }

        public long nextLong() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Long.parseLong(more);
        }

        public double nextDouble() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Double.parseDouble(more);
        }

        public String nextString(){
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return more;
        }

        public char nextChar(){
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return more.charAt(0);
        }
    }
}

Java的确没有C++方便,少了操作符重载和Pair只能靠自己写函数,显得代码很冗长。

本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。
原始发表:2017-09-28 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体同步曝光计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
目录
  • 挑战程序竞赛系列(86):3.6极限情况(3)
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档