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(92):3.6凸包(3)

挑战程序竞赛系列(92):3.6凸包(3)

作者头像
用户1147447
发布2018-01-02 10:52:44
5710
发布2018-01-02 10:52:44
举报
文章被收录于专栏:机器学习入门机器学习入门

挑战程序竞赛系列(92):3.6凸包(3)

传送门:POJ 1912: A highway and the seven dwarfs

题意:

高铁与七个小矮人:侏儒岛上有N栋房子,组成一个社区。现给定一条高铁,问该高铁是否分割了社区?

alt text
alt text

此题不错,好题。很明显,判断社区是否被分割时,只需判断外围的点集即可(凸包上的点),因为凸包所能围成的区域最大,那么社区更容易被分割。

求完凸包上的点后,朴素的做法,暴力枚举每个点对,判断是否与直线相交,肯定超时,不写了,也没啥技术含量。

实际上,凸包上的很多点也是冗余的,对直线相交判断没有任何贡献,如何说?

看图:

alt text
alt text

实际上,给定一条直线,我们还能求出凸包中关于这条直线的最大间距,真正的目的在于max d,那么什么时候能够max d呢?不难想象,对应的凸包上的极角分别旋转一点即可,平行的时候达到极值。对应图中,我们要找到A2点和A4点。

而凸包在枚举每个点对的极角时,是自然有序的,但需要做一些处理,因为在求解A4A1对应的极角时,为负值需要加个2PI,当然A1A2也是负值,却不需要加,因为凸包的起点是按照x轴排序的,注意下细节就好。

代码如下:

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.Arrays;
import java.util.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201709/P1912.txt";

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

    static final int MAX_N = 100000 + 16;
    static final double EPS = 1E-10;
    static final double PI  = Math.acos(-1.0);

    class P implements Comparable<P>{

        double x;
        double y;

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

        double add(double a, double b) {
            if (Math.abs(a + b) < EPS * (Math.abs(a) + Math.abs(b))) return 0;
            return a + b;   
        }

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

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

        double dot(P a) {
            return a.x * x + a.y * y;
        }

        double det(P a) {
            return x * a.y - y * a.x;
        }

        @Override
        public int compareTo(P o) {
            int cmp = Double.compare(x, o.x);
            return cmp != 0 ? cmp : Double.compare(y, o.y);
        }
    }

    int N;
    P[] ps;

    P[] convexHull() {
        Arrays.sort(ps);
        if (N <= 1) return ps;

        P[] qs = new P[N * 2];
        int k = 0;
        for (int i = 0; i < N; qs[k++] = ps[i++]) {
            while (k > 1 && qs[k - 1].sub(qs[k - 2]).det(ps[i].sub(qs[k - 1])) < EPS) k--;
        }

        for (int i = N - 2, t = k; i >= 0; qs[k++] = ps[i--]) {
            while (k > t && qs[k - 1].sub(qs[k - 2]).det(ps[i].sub(qs[k - 1])) < EPS) k--;
        }

        P[] res = new P[k - 1];
        System.arraycopy(qs, 0, res, 0, k - 1);
        return res;
    }

    double angle(P a, P b) {
        P t = b.sub(a);
        double ang = Math.atan2(t.y, t.x);
        if (ang < -PI / 2 + EPS) ang += 2 * PI;
        return ang;
    }

    int lowerBound(double[] ds, double v) {
        int l = 0, r = ds.length;
        while (l < r) {
            int m = (r + l) >> 1;
            if (ds[m] < v) l = m + 1;
            else r = m;
        }
        return l;
    }

    //a, b是否在直线st的两侧
    boolean check(P a, P b, P s, P t) {
        return a.sub(s).det(t.sub(s)) * b.sub(s).det(t.sub(s)) <= 0;
    }

    boolean cross(P[] qs, double[] ds, P p1, P p2) {
        int i = lowerBound(ds, angle(p1, p2));
        int j = lowerBound(ds, angle(p2, p1));
        i %= qs.length;
        j %= qs.length;

        return check(qs[i], qs[j], p1, p2);
    }

    void solve() {

        P[] qs = convexHull();
        int n = qs.length;
        double[] ds = new double[n];

        for (int i = 0; i < n; ++i) {
            ds[i] = angle(qs[i], qs[(i + 1) % n]);
        }

        while (more()) {
            P p1 = new P(nd(), nd());
            P p2 = new P(nd(), nd());
            if (N <= 1 || !cross(qs, ds, p1, p2)) out.println("GOOD");
            else out.println("BAD");
        }
    }

    void read() {
        N = ni();
        ps = new P[N];
        for (int i = 0; i < N; ++i) {
            ps[i] = new P(nd(), nd());
        }
        solve();
    }

    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);
        }
    }
}
alt text
alt text
本文参与 腾讯云自媒体分享计划,分享自作者个人站点/博客。
原始发表:2017-09-29 ,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 腾讯云自媒体分享计划  ,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(92):3.6凸包(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