Solving with Ruby and Java AtCoder ABC129 D 2D array

Introduction

• AtCoder Problems * Use Recommendations to solve past problems. Thanks to AtCoder and AtCoder Problems.

This theme

AtCoder Beginner Contest D - Lamp Difficulty: 1080

This theme, 2D array

What you are doing is simple.

Scan a total of 2 times to find the range of light in the left-right direction. Next, scan up and down twice to find the range of light in the up and down direction.

Sum the values in the left and right array and the top and bottom array to find the maximum value. Java

lamp.java

import java.util.*;

class Main {
    public static void main(String[] args) {
        final Scanner sc = new Scanner(System.in);
        final int H = Integer.parseInt(sc.next());
        final int W = Integer.parseInt(sc.next());
        final char S[][] = new char[H+2][W+2];
        for (int i=1; i<H+1; i++) {
            S[i] =  ("#" + sc.next() + "#").toCharArray();
        }
        sc.close();
        for (int i=0; i<W+2; i++) {
            S[0][i] = '#';
            S[H+1][i] = '#';
        }
        int lr[][] = new int[H+2][W+2];
        int ud[][] = new int[H+2][W+2];

        for (int i=1; i<H+1; i++) {
            int cnt = 0;
            for (int j=0; j<W+1; j++) {
                if (S[i][j]=='.') {
                    cnt++;
                } else {
                    cnt = 0;
                }
                lr[i][j] = cnt;
            }
            for (int j=W; j>0; j--) {
                if (lr[i][j]==0) {
                    cnt = 0;
                } else if (cnt==0) {
                    cnt = lr[i][j];
                } else {
                    lr[i][j] = cnt;
                }
            }
        }
        for (int j=1; j<W+1; j++) {
            int cnt = 0;
            for (int i=0; i<H+1; i++) {
                if (S[i][j]=='.') {
                    cnt++;
                } else {
                    cnt = 0;
                }
                ud[i][j] = cnt;
            }
            for (int i=H; i>0; i--) {
                if (ud[i][j]==0) {
                    cnt = 0;
                } else if (cnt==0) {
                    cnt = ud[i][j];
                } else {
                    ud[i][j] = cnt;
                }
            }
        }

        int ans = 0;
        for (int i=1; i<H+1; i++) {
            for (int j=1; j<W+1; j++) {
                int cnt = lr[i][j] + ud[i][j];
                if (ans<cnt) ans = cnt;
            }
        }
        ans -= 1;
        System.out.println(ans);
    }
}

It's a cord without any twist.

About competing in script languages etc.

• About competing in scripting languages *

A blog about the speed of scripting languages. Among them, the one that is taken up as a strict example in the script language is this * D --Lamp *.

Ruby

ruby.rb

h, w = gets.split.map(&:to_i)
s = Array.new(h + 1).map{Array.new(w + 1, 0)}
1.upto(h) do |i|
  c = 0
  l = 1
  b = gets
  1.upto(w) do |j|
    if b[j - 1] == '.'
      l = j if c == 0
      c += 1
    elsif c > 0
      l.upto(j - 1) do |k|
        s[i][k] = c
      end
      c = 0
    end
    if j == w && c > 0
      l.upto(j) do |k|
        s[i][k] = c
      end
    end
  end
end
ans = 0
1.upto(w) do |j|
  c = 0
  l = 1
  1.upto(h) do |i|
    if (s[i][j] > 0)
      l = i if c == 0
      c += 1
    elsif c > 0
      l.upto(i - 1) do |k|
        ans = s[k][j] + c if ans < s[k][j] + c
      end
      c = 0
    end
    if i == h && c > 0
      l.upto(i) do |k|
        ans = s[k][j] + c if ans < s[k][j] + c
      end
    end
  end
end
puts ans - 1

It was TLE in the old environmentRuby (2.3.3) , but it seems to pass somehow in the new environmentRuby (2.7.1) .

How fast

Roughly, it seems to be 1.4 times faster. The era of ruby has arrived.

Summary

• ABC 129 D passed
• Ruby is faster
• It's not as good as a compilation system

Referenced site

• About competing in scripting languages *