leetcode 130: Surrounded Regions

The algorithm is not so hard to think. Just try from the boarder, if there is an 'O', DFS or BFS and change all 'O's in that region to another character. After that, change all 'O's to 'X' and all different characters (I use 'n') to 'O'. The trick when using
BFS here is to change the character before push into the queue, or it will push some positions that are already in the queue and finally cause TLE.

class Solution {
public:
void solve(vector<vector<char>>& board) {
if(board.empty())
return;
int m=board.size();
int n=board[0].size();
for(int i=0;i<m;i++)
{
if(board[i][0]=='O')
bfs(board,i,0,m,n);
if(board[i][n-1]=='O')
bfs(board,i,n-1,m,n);
}
for(int i=0;i<n;i++)
{
if(board[0][i]=='O')
bfs(board,0,i,m,n);
if(board[m-1][i]=='O')
bfs(board,m-1,i,m,n);
}
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
if(board[i][j]=='O')
board[i][j]='X';
else if(board[i][j]=='n')
board[i][j]='O';
}
}
void bfs(vector<vector<char> >& board,int r,int c,int m,int n)
{
queue<pair<int,int> > q;
board[r][c]='n';
q.push(make_pair(r,c));
while(!q.empty())
{
r=q.front().first;
c=q.front().second;
q.pop();
if(r>0&&board[r-1][c]=='O')
{
board[r-1][c]='n';
q.push(make_pair(r-1,c));
}
if(r<m-1&&board[r+1][c]=='O')
{
board[r+1][c]='n';
q.push(make_pair(r+1,c));
}
if(c>0&&board[r][c-1]=='O')
{
board[r][c-1]='n';
q.push(make_pair(r,c-1));
}
if(c<n-1&&board[r][c+1]=='O')
{
board[r][c+1]='n';
q.push(make_pair(r,c+1));
}
}
}
};