经典面试题--寻找01矩阵中最大的1矩形(POJ 3494) - - 博客频道 - CSDN.NET


Leetcode 85 - Maximal Rectangle
POJ地址:http://poj.org/problem?id=3494
Given a m-by-n (0,1)-matrix, of all its submatrices of all 1’s which is the largest? By largest we mean that the submatrix has the most elements.

题目简述

题目的描述很简单,在一个M * N的矩阵中,所有的元素只有0和1, 找出只包含1的最大矩形。
例如:图中是一个4 × 6的矩形,画出红色的是我们要找到的区域。




这样的话,其实我们要找到的某个矩形就转换成 一某一个行开始的 histogram的最大矩形问题了。
那么我们原始矩形可以变成如下的形式的数据:
第一行表示,我们以第一行作为底边,所形成的 histogram的高度,其他行也类似。
所以问题变成了 枚举每一行,然后求出每一行对应的histogram的最大矩形。
关于histogram求最大矩形
  1. int nRow, nCol;  
  2. int matrix[MAX_LEN][MAX_LEN]; //原数据  
  3. int heights[MAX_LEN][MAX_LEN];//用这个数组来描述 histogram,其中heights[i]表示 以第i行走底的histogram,里面的元素表示对应列的高度  
  4. struct Node  
  5. {  
  6.     int height;  
  7.     int position;  
  8.     Node(int _height, int _from): height(_height), position(_from)  
  9.     {  }  
  10.     Node()  
  11.     {  }  
  12. };  
  13. int max(int a, int b)  
  14. {  
  15.     return a>b ? a : b;  
  16. }  
  17. int GetArea(int iRow) //用单调栈来枚举其中以 某一行做底的 histogram 所得到的最大矩形面积。  
  18. {  
  19.     topID = 0;  
  20.     push(Node(-1, 0)); 
  21.     int i;  
  22.     int area, maxArea = 0;  
  23.     int position, height;
  24.     for( i = 0; i <= nCol; i++)  
  25.     {  
  26.         position = i + 1;  
  27.         if(i == nCol)  
  28.         {  
  29.             height = -1;  
  30.         }  
  31.         else  
  32.         {  
  33.             height = heights[iRow][i];  
  34.         }  
  35.         Node t(height, position);  
  36.         while(top().height > height)  
  37.         {  
  38.             t = top();  
  39.             pop();  
  40.             area = (position - t.position) * t.height;  
  41.             if(area > maxArea)  
  42.             {  
  43.                 maxArea = area;  
  44.             }  
  45.         }  
  46.         push(Node(height, t.position));  
  47.     }  
  48.     return maxArea;  
  49. }  
  50. int main()  
  51. {  
  52.     while(scanf("%d%d", &nRow, &nCol) != EOF)  
  53.     {  
  54.         int i,j;  
  55.         int b;  
  56.         for( i = 0;i < nRow; i++)  
  57.         {  
  58.             for( j = 0; j < nCol; j++)  
  59.             {  
  60.                 scanf("%d", &matrix[i][j]);   
  61.             }  
  62.         }  
  63.                 //求histogram,求的时候,如果以 i 行为底边,j对应的高度是 从i 到 最高连续的1 的数量  
  64.         memcpy(heights, matrix, sizeof(matrix));  
  65.         for( i = 0; i < nCol; i++)  
  66.         {  
  67.             for( j = 1; j < nRow; j++)  
  68.             {  
  69.                 if(heights[j][i] != 0)  
  70.                 {  
  71.                     heights[j][i] += heights[j-1][i];  
  72.                 }  
  73.             }  
  74.         }  
  75.         int maxArea = 0, Area;  
  76.         for( i = 0; i < nRow; i++)  
  77.         {  
  78.             Area = GetArea(i);  
  79.             maxArea = max(maxArea, Area);  
  80.         }  
  81.         printf("%d\n", maxArea);  
  82.     }   
  83. }  
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