## Surrounded Regions

### 描述

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


### 代码

// LeetCode, Surrounded Regions
// BFS，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
void solve(vector<vector<char>> &board) {
if (board.empty()) return;

const int m = board.size();
const int n = board[0].size();
for (int i = 0; i < n; i++) {
bfs(board, 0, i);
bfs(board, m - 1, i);
}
for (int j = 1; j < m - 1; j++) {
bfs(board, j, 0);
bfs(board, j, n - 1);
}
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] == '+')
board[i][j] = 'O';
}
private:
void bfs(vector<vector<char>> &board, int i, int j) {
typedef pair<int, int> state_t;
queue<state_t> q;
const int m = board.size();
const int n = board[0].size();

auto state_is_valid = [&](const state_t &s) {
const int x = s.first;
const int y = s.second;
if (x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'O')
return false;
return true;
};

auto state_extend = [&](const state_t &s) {
vector<state_t> result;
const int x = s.first;
const int y = s.second;
// 上下左右
const state_t new_states[4] = {{x-1,y}, {x+1,y},
{x,y-1}, {x,y+1}};
for (int k = 0; k < 4;  ++k) {
if (state_is_valid(new_states[k])) {
// 既有标记功能又有去重功能
board[new_states[k].first][new_states[k].second] = '+';
result.push_back(new_states[k]);
}
}

return result;
};

state_t start = { i, j };
if (state_is_valid(start)) {
board[i][j] = '+';
q.push(start);
}
while (!q.empty()) {
auto cur = q.front();
q.pop();
auto new_states = state_extend(cur);
for (auto s : new_states) q.push(s);
}
}
};