Puzzle: găsiți cel mai mare dreptunghi (problema dreptunghiulară maximă)

Iată o pagină care are un cod și o analiză.

Problema dvs. specială începe puțin în pagină, căutați pagina pentru problema dreptunghiulară maximă .

http://www.seas.gwu.edu/~ simhaweb / cs151 / prelegeri / module6 / module6.html

@lassevk

    // 4. Outer double-for-loop to consider all possible positions 
    //    for topleft corner. 
    for (int i=0; i < M; i++) {
      for (int j=0; j < N; j++) {

        // 2.1 With (i,j) as topleft, consider all possible bottom-right corners. 

        for (int a=i; a < M; a++) {
          for (int b=j; b < N; b++) {

HAHA ... O (m2 n2) .. Asta probabil aș fi venit.

Văd că ei continuă să dezvolte optmizări ... arată bine, o să citesc.

@lassevk

I found the referenced article, from DDJ: The Maximal Rectangle Problem

Dupa ce am luptat atat de mult am scris acest algoritm ... Vedeti codul ...

Înțelegi asta.Acest cod vorbește!

import java.util.Scanner;
import java.util.Stack;

/**
 * Created by BK on 05-08-2017.
 */
public class LargestRectangleUnderHistogram {
    public static void main(String... args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] input = new int[n];
        for (int j = 0; j < n; j++) {
            input[j] = scanner.nextInt();
        }



        /*
        *   This is the procedure used for solving :
        *
        *   Travel from first element to last element of the array
        *
        *   If stack is empty add current element to stack
        *
        *   If stack is not empty check for the top element of stack if
        *   it is smaller than the current element push into stack
        *
        *   If it is larger than the current element pop the stack until we get an
        *   element smaller than the current element or until stack becomes empty
        *
        *   After popping each element check if the stack is empty or not..
        *
        *   If stack is empty it means that this is the smallest element encountered till now
        *
        *   So we can multiply i with this element to get a big rectangle which is contributed by
        *
        *   this
        *
        *   If stack is not empty then check for individual areas(Not just one bar individual area means largest rectangle by this) (i-top)*input[top]
        *
        *
        * */

        /*
        * Initializing the maxarea as we check each area with maxarea
        */

        int maxarea = -1;
        int i = 0;
        Stack stack = new Stack<>();
        for (i = 0; i < input.length; i++) {

            /*
            *   Pushing the element if stack is empty
            * */


            if (stack.isEmpty()) {
                stack.push(i);
            } else {

                /*
                *   If stack top element is less than current element push
                * */

                if (input[stack.peek()] < input[i]) {
                    stack.push(i);
                } else {

                    /*
                    *   Else pop until we encounter an element smaller than this in stack or stack becomes empty
                    *   
                    * */


                    while (!stack.isEmpty() && input[stack.peek()] > input[i]) {

                        int top = stack.pop();

                        /*
                        *   If stack is empty means this is the smallest element encountered so far
                        *   
                        *   So we can multiply this with i
                        * */

                        if (stack.isEmpty()) {
                            maxarea = maxarea < (input[top] * i) ? (input[top] * i) : maxarea;
                        }

                        /*
                         *  If stack is not empty we find areas of each individual rectangle
                         *  Remember we are in while loop
                         */

                        else {
                            maxarea = maxarea < (input[top] * (i - top)) ? (input[top] * (i - top)) : maxarea;
                        }
                    }
                    /*
                    *   Finally pushing the current element to stack
                    * */

                    stack.push(i);
                }
            }
        }

        /*
        *  This is for checking if stack is not empty after looping the last element of input
        *  
        *  This happens if input is like this 4 5 6 1 2 3 4 5
        *  
        *  Here 2 3 4 5 remains in stack as they are always increasing and we never got 
        *  
        *  a chance to pop them from stack by above process
        *  
        * */


        while (!stack.isEmpty()) {

            int top = stack.pop();

            maxarea = maxarea < (i - top) * input[top] ? (i - top) * input[top] : maxarea;
        }

        System.out.println(maxarea);
    }
}

Am implementat soluția Dobbs în Java.

Nicio garanție pentru nimic.

package com.test;

import java.util.Stack;

public class Test {

    public static void main(String[] args) {
        boolean[][] test2 = new boolean[][] { new boolean[] { false, true, true, false },
                new boolean[] { false, true, true, false }, new boolean[] { false, true, true, false },
                new boolean[] { false, true, false, false } };
        solution(test2);
    }

    private static class Point {
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }

        public int x;
        public int y;
    }

    public static int[] updateCache(int[] cache, boolean[] matrixRow, int MaxX) {
        for (int m = 0; m < MaxX; m++) {
            if (!matrixRow[m]) {
                cache[m] = 0;
            } else {
                cache[m]++;
            }
        }
        return cache;
    }

    public static void solution(boolean[][] matrix) {
        Point best_ll = new Point(0, 0);
        Point best_ur = new Point(-1, -1);
        int best_area = 0;

        final int MaxX = matrix[0].length;
        final int MaxY = matrix.length;

        Stack stack = new Stack();
        int[] cache = new int[MaxX + 1];

        for (int m = 0; m != MaxX + 1; m++) {
            cache[m] = 0;
        }

        for (int n = 0; n != MaxY; n++) {
            int openWidth = 0;
            cache = updateCache(cache, matrix[n], MaxX);
            for (int m = 0; m != MaxX + 1; m++) {
                if (cache[m] > openWidth) {
                    stack.push(new Point(m, openWidth));
                    openWidth = cache[m];
                } else if (cache[m] < openWidth) {
                    int area;
                    Point p;
                    do {
                        p = stack.pop();
                        area = openWidth * (m - p.x);
                        if (area > best_area) {
                            best_area = area;
                            best_ll.x = p.x;
                            best_ll.y = n;
                            best_ur.x = m - 1;
                            best_ur.y = n - openWidth + 1;
                        }
                        openWidth = p.y;
                    } while (cache[m] < openWidth);
                    openWidth = cache[m];
                    if (openWidth != 0) {
                        stack.push(p);
                    }
                }
            }
        }

        System.out.printf("The maximal rectangle has area %d.\n", best_area);
        System.out.printf("Location: [col=%d, row=%d] to [col=%d, row=%d]\n", best_ll.x + 1, best_ll.y + 1,
                best_ur.x + 1, best_ur.y + 1);
    }

}

Eu sunt autorul articolului dr. Dobb și am întrebat ocazional despre o implementare. Aici este un simplu în C:

#include 
#include 
#include 

typedef struct {
  int one;
  int two;
} Pair;

Pair best_ll = { 0, 0 };
Pair best_ur = { -1, -1 };
int best_area = 0;

int *c; /* Cache */
Pair *s; /* Stack */
int top = 0; /* Top of stack */

void push(int a, int b) {
  s[top].one = a;
  s[top].two = b;
  ++top;
}

void pop(int *a, int *b) {
  --top;
  *a = s[top].one;
  *b = s[top].two;
}


int M, N; /* Dimension of input; M is length of a row. */

void update_cache() {
  int m;
  char b;
  for (m = 0; m!=M; ++m) {
    scanf(" %c", &b);
    fprintf(stderr, " %c", b);
    if (b=='0') {
      c[m] = 0;
    } else { ++c[m]; }
  }
  fprintf(stderr, "\n");
}


int main() {
  int m, n;
  scanf("%d %d", &M, &N);
  fprintf(stderr, "Reading %dx%d array (1 row == %d elements)\n", M, N, M);
  c = (int*)malloc((M+1)*sizeof(int));
  s = (Pair*)malloc((M+1)*sizeof(Pair));
  for (m = 0; m!=M+1; ++m) { c[m] = s[m].one = s[m].two = 0; }
  /* Main algorithm: */
  for (n = 0; n!=N; ++n) {
    int open_width = 0;
    update_cache();
    for (m = 0; m!=M+1; ++m) {
      if (c[m]>open_width) { /* Open new rectangle? */
        push(m, open_width);
        open_width = c[m];
      } else /* "else" optional here */
      if (c[m]best_area) {
            best_area = area;
            best_ll.one = m0; best_ll.two = n;
            best_ur.one = m-1; best_ur.two = n-open_width+1;
          }
          open_width = w0;
        } while (c[m]

16 12
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0
0 0 0 1 1 0 0 1 0 0 0 1 1 0 1 0
0 0 0 1 1 0 1 1 1 0 1 1 1 0 1 0
0 0 0 0 1 1 * * * * * * 0 0 1 0
0 0 0 0 0 0 * * * * * * 0 0 1 0
0 0 0 0 0 0 1 1 0 1 1 1 1 1 1 0
0 0 1 0 0 0 0 1 0 0 1 1 1 0 1 0 
0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0

The maximal rectangle has area 12.
Location: [col=7, row=6] to [col=12, row=5]