《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：

  1 // 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.

  2 // I'll return the size and the left top corner of the subsquare.

  3 #include <iostream>

  4 #include <vector>

  5 using namespace std;

  6 

  7 class Solution {

  8 public:

  9     void maxSubsquare(const vector<vector<int> > &matrix, int &max_left, int &max_top, int &max_size) {

 10         int n = matrix.size();

 11         

 12         max_left = max_top = max_size = -1;

 13         

 14         if (n <= 1) {

 15             return;

 16         }

 17         

 18         vector<vector<int> > top   (n, vector<int>(n));

 19         vector<vector<int> > bottom(n, vector<int>(n));

 20         vector<vector<int> > left  (n, vector<int>(n));

 21         vector<vector<int> > right (n, vector<int>(n));

 22         

 23         int i, j;

 24         int tmp;

 25         

 26         // use DP to preprocess the data, count how many consecutive 1s are there to the left, right, top, bottom of matrix[i][j].

 27         for (i = 0; i <= n - 1; ++i) {

 28             tmp = 0;

 29             for (j = 0; j <= n - 1; ++j) {

 30                 left[i][j] = matrix[i][j] ? (++tmp) : (tmp = 0);

 31             }

 32         }

 33         for (j = 0; j <= n - 1; ++j) {

 34             tmp = 0;

 35             for (i = 0; i <= n - 1; ++i) {

 36                 top[i][j] = matrix[i][j] ? (++tmp) : (tmp = 0);

 37             }

 38         }

 39         for (i = n - 1; i >= 0; --i) {

 40             tmp = 0;

 41             for (j = n - 1; j >= 0; --j) {

 42                 right[i][j] = matrix[i][j] ? (++tmp) : (tmp = 0);

 43             }

 44         }

 45         for (j = n - 1; j >= 0; --j) {

 46             tmp = 0;

 47             for (i = n - 1; i >= 0; --i) {

 48                 bottom[i][j] = matrix[i][j] ? (++tmp) : (tmp = 0);

 49             }

 50         }

 51         

 52         int len;

 53         // O(n ^ 3) solution with O(n ^ 2) space usage.

 54         for (i = 0; i < n; ++i) {

 55             for (j = 0; j < n; ++j) {

 56                 for (len = 2; len + i <= n && len + j <= n; ++len) {

 57                     if (right[i][j] < len || bottom[i][j] < len) {

 58                         continue;

 59                     }

 60                     if (left[i][j + len - 1] < len || bottom[i][j + len - 1] < len) {

 61                         continue;

 62                     }

 63                     if (right[i + len - 1][j] < len || top[i + len - 1][j] < len) {

 64                         continue;

 65                     }

 66                     if (left[i + len - 1][j + len - 1] < len || top[i + len - 1][j + len - 1] < len) {

 67                         continue;

 68                     }

 69                     // all four borders are '1's.

 70                     if (len > max_size) {

 71                         max_top = i;

 72                         max_left = j;

 73                         max_size = len;

 74                     }

 75                 }

 76             }

 77         }

 78         

 79         // clear up data

 80         for (i = 0; i < n; ++i) {

 81             left[i].clear();

 82             right[i].clear();

 83             top[i].clear();

 84             bottom[i].clear();

 85         }

 86         left.clear();

 87         right.clear();

 88         top.clear();

 89         bottom.clear();

 90     };

 91 };

 92 

 93 int main()

 94 {

 95     int n;

 96     int i, j;

 97     vector<vector<int> > matrix;

 98     Solution sol;

 99     int max_left, max_top, max_size;

100     

101     while (cin >> n && n > 0) {

102         matrix.resize(n);

103         for (i = 0; i < n; ++i) {

104             matrix[i].resize(n);

105         }

106         

107         for (i = 0; i < n; ++i) {

108             for (j = 0; j < n; ++j) {

109                 cin >> matrix[i][j];

110             }

111         }

112         

113         sol.maxSubsquare(matrix, max_left, max_top, max_size);

114         if (max_size > 0) {

115             cout << max_top << ' ' << max_left << endl;

116             cout << max_top << ' ' << max_left + max_size - 1 << endl;

117             cout << max_top + max_size - 1 << ' ' << max_left << endl;

118             cout << max_top + max_size - 1 << ' ' << max_left + max_size - 1 << endl;

119         } else {

120             cout << "No subsquare found." << endl;

121         }

122         

123         for (i = 0; i < n; ++i) {

124             matrix[i].clear();

125         }

126         matrix.clear();

127     }

128     

129     return 0;

130 }