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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)

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发布2018-01-02 10:52:44
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发布2018-01-02 10:52:44
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文章被收录于专栏: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?

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:

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:

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.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);
        }
    }
}
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