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

  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 }