Leetcode Surrounded Regions

class Solution {
public:
    void solve(vector<vector<char> > &board) {
        if (board.size() == 0) return;
        int my = board.size() - 1;
        int mx = board[0].size() -1;
        // top border
        for (int i=0; i<=mx; i++) {
            bfs(board, mx, my, i, 0, 'O', '+');
        }
        // bottom border
        for (int i=0; i<=mx; i++) {
            bfs(board, mx, my, i, my, 'O', '+');
        }
        // left border
        for (int i=0; i<=my; i++) {
            bfs(board, mx, my, 0, i, 'O', '+');
        }
        // right border
        for (int i=0; i<=my; i++) {
            bfs(board, mx, my, mx, i, 'O', '+');
        }
        
        // scan & fill
        for (int i=1; i<=my-1; i++) {
            for (int j=1; j<=mx-1; j++) {
                if (board[i][j] == 'O') board[i][j] = 'X';
            }
        }
        
        // restore
        for (int i=0; i<=my; i++) {
            for (int j=0; j<=mx; j++) {
                if (board[i][j] == '+') board[i][j] = 'O';
            }
        }
    }
    
    void bfs(vector<vector<char> >& board, int mx, int my, int _x, int _y, char oc, char nc) {
        queue<pair<int, int> > que;
        que.push(pair<int, int>(_x, _y));
        
        while (!que.empty()) {
            pair<int, int> blk = que.front();
            que.pop();
            int x = blk.first;
            int y = blk.second;
            if (y <= my && x <= mx && board[y][x] == oc) {
                board[y][x] = nc;
                if (x+1 <= mx && board[y][x+1] == oc) que.push(pair<int, int>(x+1, y));
                if (x-1 >= 0 && board[y][x-1] == oc) que.push(pair<int, int>(x-1, y));
                if (y+1 <= my && board[y+1][x] == oc) que.push(pair<int, int>(x, y+1));
                if (y-1 >= 0 && board[y-1][x] == oc) que.push(pair<int, int>(x, y-1));
            }
        }
    }
};

题目给出的输入输出样例（上输入，下输出）

X X X X
X O O X
X X O X
X O X X

X X X X
X X X X
X X X X
X O X X

由题目给出的例子可以看出与边界上的'O'联通的都不能算作被'X'包围，那么我们先把这些点标记一下，然后扫描整个board将未标记且为'O'的元素设置为'X'即可。图形学上应该把这个叫做种子填充算法，当然这种朴素的算法效率在像素大的情况下速度肯定会不能忍受，于是有了扫描线填充（与多边形扫描线填充一样也利用了局部的连续性）等改进算法。不过这里时间要求不苛刻，也没必要实现了。

参考：

zhuli哥的题解 http://www.cnblogs.com/zhuli19901106/p/3570554.html

Flood_fill http://en.wikipedia.org/wiki/Flood_fill