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社区首页 >专栏 >挑战程序竞赛系列(91):3.6凸包(2)

挑战程序竞赛系列(91):3.6凸包(2)

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用户1147447
发布2018-01-02 10:53:00
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发布2018-01-02 10:53:00
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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(91):3.6凸包(2)

传送门:POJ 1113: Wall

题意参考hankcs: http://www.hankcs.com/program/algorithm/poj-1113-wall.html

题意:

墙:有个愚蠢的皇帝要你造墙将城堡围起来,城堡的顶点有N个,墙必须离城堡的边至少L单位远,并且墙的总长度尽量小。求此长度?

所以求最大凸包把所有的点都围起来即可,但是为了保证围墙的距离最少,拐点用半径为L的圆包围起来,此题巧妙之处在于,凸包的所有踵点的周长实际就是一个圆的周长。

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/P1113.txt";

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

    static final int MAX_N = 1000 + 16;
    static final double PI = Math.acos(-1.0);

    class P implements Comparable<P>{
        int x;
        int y;

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

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

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

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

        @Override
        public int compareTo(P o) {
            return this.x != o.x ? this.x - o.x : this.y - o.y;
        }
    }

    int N, L;
    P[] ps;

    double dist(P a, P b) {
        return Math.sqrt(a.sub(b).dot(a.sub(b)));
    }

    P[] convexHull() {

        Arrays.sort(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])) <= 0) 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])) <= 0) k --;
        }

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

    void solve() {
        P[] qs = convexHull();
        double res = 0;
        int n = qs.length;
        for (int i = 0; i < n; ++i) {
            res += dist(qs[i], qs[(i + 1) % n]);
        }

        out.printf("%.0f\n", 2 * PI * L + res);
    }


    void read() {
        N = ni();
        L = ni();

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

        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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原始发表:2017-09-29 ,如有侵权请联系 cloudcommunity@tencent.com 删除

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