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(88):3.6平面扫描(2)

挑战程序竞赛系列(88):3.6平面扫描(2)

作者头像
用户1147447
发布2018-01-02 10:48:28
5060
发布2018-01-02 10:48:28
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(88):3.6平面扫描(2)

传送门:POJ 3168: Barn Expansion

题意:

求不重叠的矩形个数。

当然,题意是说求能够被扩大的矩形个数,很坑爹,一开始以为扩大的单位是1格,其实题目中明确指出: at least some amount without bumping into another barn,英文是硬伤,所以只要不重叠即可。

思路参考hankcs,把矩形抽象成一个个区间,对应于水平区间和垂直区间,定义每个区间的起点和终点。设置一个计数器,遇到起点计数+1,遇到终点计数-1。

那么矩形不重叠等价于对应于的起点计数不超过2。

这里写图片描述
这里写图片描述

接着分别沿x轴扫描,和y轴扫描即可。注意一个细节,排序时,横纵坐标相同时,起点优先,因为corner相碰算重叠。

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/P3168.txt";

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

    static final int MAX_N = 25000 + 16;

    class Point implements Comparable<Point>{
        int ord;         // 当前轴坐标
        int pos;         // 另一轴上的坐标
        int id;          // 对应的矩阵id
        int type;        // 起点: 0 终点: 1

        Point(int ord, int pos, int id, int type){
            this.ord  = ord;
            this.pos  = pos;
            this.id   = id;
            this.type = type;
        }

        @Override
        public int compareTo(Point that) {
            if (this.ord != that.ord) {
                return this.ord - that.ord;
            }
            if (this.pos != that.pos) {
                return this.pos - that.pos;
            }
            return this.type - that.type;
        }
    }

    int N;
    Point[] toX = new Point[4 * MAX_N];     // 以x轴方向扫描
    Point[] toY = new Point[4 * MAX_N];     // 以y轴方向扫描
    boolean[] ok = new boolean[MAX_N];

    void scan(Point[] ps) {

        for (int i = 0; i < 4 * N; ++i) {

            int j = i;
            while (j < 4 * N && ps[j].ord == ps[i].ord) j++;

            boolean invalid = false;
            int share = 0;

            for (int k = i; k < j; ++k) {
                Point p = ps[k];
                if (invalid) {
                    ok[p.id] = false;
                }
                if (p.type == 0) {
                    share ++;
                    if (share >= 2) {
                        invalid = true;
                    }
                }
                else {
                    share --;
                    if (share == 0) {
                        invalid = false;
                    }
                }
            }
            i = j - 1;
        }
    }

    void solve() {

        Arrays.sort(toX, 0, 4 * N);
        Arrays.sort(toY, 0, 4 * N);

        Arrays.fill(ok, true);

        scan(toX);
        scan(toY);

        int ans = 0;
        for (int i = 0; i < N; ++i) {
            if (ok[i]) ans ++;
        }

        out.println(ans);
    }

    void read() {
        N = ni();
        for (int i = 0; i < N; ++i) {
            int A = ni(), B = ni(), C = ni(), D = ni();
            toX[4 * i]     = new Point(A, B, i, 0);
            toX[4 * i + 1] = new Point(A, D, i, 1);
            toX[4 * i + 2] = new Point(C, B, i, 0);
            toX[4 * i + 3] = new Point(C, D, i, 1);

            toY[4 * i]     = new Point(B, A, i, 0);
            toY[4 * i + 1] = new Point(B, C, i, 1);
            toY[4 * i + 2] = new Point(D, A, i, 0);
            toY[4 * i + 3] = new Point(D, C, i, 1);
        }

        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(88):3.6平面扫描(2)
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