《Cracking the Coding Interview》——第18章：难题——题目11

2014-04-29 04:30

题目：给定一个由‘0’或者‘1’构成的二维数组，找出一个四条边全部由‘1’构成的正方形（矩形中间可以有‘0’），使得矩形面积最大。

解法：用动态规划思想，记录二维数组每个元素向上下左右四个方向各有多少个连续的‘1’，然后用O(n^3)时间计算出满足条件的最大正方形。时间复杂度O(n^3)，空间复杂度O(n^2)。

代码：

 // 18.11 Given an NxN matrix of 0s and 1s, find out a subsquare whose all four borders are all 1s. If multiple satisfies the condition, any one is OK.
 // I'll return the size and the left top corner of the subsquare.
 #include <iostream>
 #include <vector>
 using namespace std;

 class Solution {
 public:
     void maxSubsquare(const vector<vector<int> > &matrix, int &max_left, int &max_top, int &max_size) {
         int n = matrix.size();

         max_left = max_top = max_size = -;

         if (n <= ) {
             return;
         }

         vector<vector<int> > top   (n, vector<int>(n));
         vector<vector<int> > bottom(n, vector<int>(n));
         vector<vector<int> > left  (n, vector<int>(n));
         vector<vector<int> > right (n, vector<int>(n));

         int i, j;
         int tmp;

         // use DP to preprocess the data, count how many consecutive 1s are there to the left, right, top, bottom of matrix[i][j].
         for (i = ; i <= n - ; ++i) {
             tmp = ;
             for (j = ; j <= n - ; ++j) {
                 left[i][j] = matrix[i][j] ? (++tmp) : (tmp = );
             }
         }
         for (j = ; j <= n - ; ++j) {
             tmp = ;
             for (i = ; i <= n - ; ++i) {
                 top[i][j] = matrix[i][j] ? (++tmp) : (tmp = );
             }
         }
         for (i = n - ; i >= ; --i) {
             tmp = ;
             for (j = n - ; j >= ; --j) {
                 right[i][j] = matrix[i][j] ? (++tmp) : (tmp = );
             }
         }
         for (j = n - ; j >= ; --j) {
             tmp = ;
             for (i = n - ; i >= ; --i) {
                 bottom[i][j] = matrix[i][j] ? (++tmp) : (tmp = );
             }
         }

         int len;
         // O(n ^ 3) solution with O(n ^ 2) space usage.
         for (i = ; i < n; ++i) {
             for (j = ; j < n; ++j) {
                 for (len = ; len + i <= n && len + j <= n; ++len) {
                     if (right[i][j] < len || bottom[i][j] < len) {
                         continue;
                     }
                     if (left[i][j + len - ] < len || bottom[i][j + len - ] < len) {
                         continue;
                     }
                     if (right[i + len - ][j] < len || top[i + len - ][j] < len) {
                         continue;
                     }
                     if (left[i + len - ][j + len - ] < len || top[i + len - ][j + len - ] < len) {
                         continue;
                     }
                     // all four borders are '1's.
                     if (len > max_size) {
                         max_top = i;
                         max_left = j;
                         max_size = len;
                     }
                 }
             }
         }

         // clear up data
         for (i = ; i < n; ++i) {
             left[i].clear();
             right[i].clear();
             top[i].clear();
             bottom[i].clear();
         }
         left.clear();
         right.clear();
         top.clear();
         bottom.clear();
     };
 };

 int main()
 {
     int n;
     int i, j;
     vector<vector<int> > matrix;
     Solution sol;
     int max_left, max_top, max_size;

     while (cin >> n && n > ) {
         matrix.resize(n);
         for (i = ; i < n; ++i) {
             matrix[i].resize(n);
         }

         for (i = ; i < n; ++i) {
             for (j = ; j < n; ++j) {
                 cin >> matrix[i][j];
             }
         }

         sol.maxSubsquare(matrix, max_left, max_top, max_size);
         if (max_size > ) {
             cout << max_top << ' ' << max_left << endl;
             cout << max_top << ' ' << max_left + max_size -  << endl;
             cout << max_top + max_size -  << ' ' << max_left << endl;
             cout << max_top + max_size -  << ' ' << max_left + max_size -  << endl;
         } else {
             cout << "No subsquare found." << endl;
         }

         for (i = ; i < n; ++i) {
             matrix[i].clear();
         }
         matrix.clear();
     }

     return ;
 }