zoukankan      html  css  js  c++  java
  • [LeetCode] Surrounded Regions

    This problem is not quite difficult (a typical BFS traversal of graph), though, its aceptance rate is relatively low. In fact, the key obstacle in passing this problem is how to reduce the number of checks and avoid the annoying TLE.

    There is a nice idea in this link. In fact, all the surrounded regions will not contain an O on the boundary. This idea takes advantage of this observation and visit the board from O's on the boundary and mark them using another character, say, #. Finally, all the remaining O's are the surrounded regions and should be captured to X. The # regions simply need to be recovered to O.

    I adopt the idea from the above link. And I include some minor optimizations. For example, I pass the queue toVisit to the update function to check and update the four neighbors together during the BFS. This turns out to reduce the running time from 28ms to 16ms.

    The code is as follows.

     1 class Solution {
     2 public:
     3     void solve(vector<vector<char>>& board) {
     4         if (board.empty()) return;
     5         int m = board.size(), n = board[0].size();
     6         for (int i = 0; i < m; i++) {
     7             if (board[i][0] == 'O')
     8                 mark(board, i, 0);
     9             if (board[i][n - 1] == 'O')
    10                 mark(board, i, n - 1);
    11         }
    12         for (int j = 0; j < n; j++) {
    13             if (board[0][j] == 'O')
    14                 mark(board, 0, j);
    15             if (board[m - 1][j] == 'O')
    16                 mark(board, m - 1, j);
    17         }
    18         capture(board);
    19     }
    20 private:
    21     // Update neighbors
    22     void update(vector<vector<char>>& board, queue<pair<int, int>>& toVisit, int r, int c) {
    23         int m = board.size(), n = board[0].size();
    24         if (r - 1 >= 0 && board[r - 1][c] == 'O') {
    25             board[r - 1][c] = '#';
    26             toVisit.push(make_pair(r - 1, c));
    27         }
    28         if (r + 1 < m && board[r + 1][c] == 'O') {
    29             board[r + 1][c] = '#';
    30             toVisit.push(make_pair(r + 1, c));
    31         }
    32         if (c - 1 >= 0 && board[r][c - 1] == 'O') {
    33             board[r][c - 1] = '#';
    34             toVisit.push(make_pair(r, c - 1));
    35         }
    36         if (c + 1 < n && board[r][c + 1] == 'O') {
    37             board[r][c + 1] = '#';
    38             toVisit.push(make_pair(r, c + 1));
    39         }
    40     }
    41     // Mark non-surrounded regions
    42     void mark(vector<vector<char>>& board, int r, int c) {
    43         queue<pair<int, int> > toVisit;
    44         toVisit.push(make_pair(r, c));
    45         board[r][c] = '#';
    46         while (!toVisit.empty()) {
    47             int num = toVisit.size();
    48             for (int i = 0; i < num; i++) {
    49                 pair<int, int> cur = toVisit.front();
    50                 toVisit.pop();
    51                 update(board, toVisit, cur.first, cur.second);
    52             }
    53         }
    54     }
    55     // Capture surrounded regions
    56     void capture(vector<vector<char>>& board) {
    57         int m = board.size(), n = board[0].size();
    58         for (int i = 0; i < m; i++) {
    59             for (int j = 0; j < n; j++) {
    60                 if (board[i][j] == '#')
    61                     board[i][j] = 'O';
    62                 else board[i][j] = 'X';
    63             }
    64         }
    65     }
    66 };
  • 相关阅读:
    【POJ 2044】 Weather Forecast
    【POJ 1703】 Find them,Catch them
    【SCOI 2005】 骑士精神
    字长与指针
    XModem协议
    SecureCRT乱码问题解决方法
    usb设备驱动程序
    如何检测 51单片机IO口的下降沿
    matlab神经网络工具箱创建神经网络
    九针串口接线问题, 232, 485
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4598456.html
Copyright © 2011-2022 走看看