leetcode 130 Surrounded Regions(BFS)

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,

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

After running your function, the board should be:

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

题解：这道题测试数据卡的很严。一开始用dfs爆栈了，后来改成bfs又超时了。。最后改进了下：先查边上有没有‘O’，和边上连着的肯定不会被围住。把这些都标记出来以后，剩下的里面的‘O’必然都要变成‘X’。

class Solution {
public:
    int n,m;
    int dx[]={,,,-};
    int dy[]={,-,,};
    vector<vector<char> >flag;

    void bfs(vector<vector<char> >& board,int x,int y){
        queue<pair<int,int> >q;
        q.push(make_pair(x,y));
        flag[x][y]=;

        while(!q.empty()){
            int qx=q.front().first;
            int qy=q.front().second;
            q.pop();
            for(int i=;i<;i++){
                int tx=qx+dx[i];
                int ty=qy+dy[i];
                if(tx>=&&tx<n&&ty>=&&ty<m&&flag[tx][ty]==&&board[tx][ty]=='O'){
                    q.push(make_pair(tx,ty));
                    flag[tx][ty]=;
                }
            }
        }
    }

    void solve(vector<vector<char> >& board) {
        if(board.empty()){
            return ;
        }
        n=board.size();
        m=board[].size();
        flag.resize(n);
        for(int i=;i<n;i++){
            flag[i].resize(m,);
        }

        for(int i=;i<m;i++){
            if(board[][i]=='O'&&flag[][i]==){
                bfs(board,,i);
            }
            if(board[n-][i]=='O'&&flag[n-][i]==){
                bfs(board,n-,i);
            }
        }

        for(int i=;i<=n-;i++){
            if(board[i][]=='O'&&flag[i][]==){
                bfs(board,i,);
            }
            if(board[i][m-]=='O'&&flag[i][m-]==){
                bfs(board,i,m-);
            }
        }

        for(int i=;i<n;i++){
            for(int j=;j<m;j++){
                if(board[i][j]=='O'&&flag[i][j]==){
                    board[i][j]='X';
                }
            }
        }
    }
};