पहेली: सबसे बड़ा आयताकार (अधिकतम आयताकार समस्या) खोजें

यहां एक पृष्ठ है जिसमें कुछ कोड और कुछ विश्लेषण हैं।

आपकी विशेष समस्या पृष्ठ पर थोड़ी सी शुरू होती है, अधिकतम आयताकार समस्या टेक्स्ट के लिए पृष्ठ खोजें।

http://www.seas.gwu.edu/~ simhaweb / cs151 / व्याख्यान / 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 ... ओ (एम 2 एन 2) .. शायद यह है कि मैं क्या आया होगा।

मुझे लगता है कि वे ऑप्टमाइजेशन विकसित करने के लिए आगे बढ़ते हैं ... अच्छा लगता है, मुझे पढ़ा जाएगा।

@lassevk

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

इतनी संघर्ष करने के बाद मैंने इस एल्गोरिदम लिखा है ... बस कोड देखें ...

तुम समझते हो.यह कोड बोलता है !!

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);
    }
}

मैंने जावा में डॉब्स के समाधान को लागू किया।

किसी भी चीज़ के लिए कोई वारंटी नहीं।

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);
    }

}

मैं डॉ डॉब के लेख का लेखक हूं और कभी-कभी कार्यान्वयन के बारे में पूछता हूं। यहां सी में एक सरल है:

#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]