Surrounded Regions

Title Description
Given an m x n matrix board containing 'X' and 'O', capture all regions that are 4-directionally surrounded by 'X'.

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

Example 1:

Input: board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
Output: [["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
Explanation: Surrounded regions should not be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.
 

Complete code：

// xxxOOO.cpp :  This file contains  "main"  function . Program execution will start and end here .
//

#include <iostream>
using namespace std;
char board[201][201];
int book[201][201];
int n, m;
void dfs(int x, int y) {
	int i, j, k,tx,ty;
	// Four directions   Right down left up 
	int next[4][2] = {
		{0,1},{1,0},{0,-1},{-1,0}
	};

	for (k = 0;k < 4;k++) {
		tx = x + next[k][0];// New coordinates 
		ty = y + next[k][1];
		// Judgement boundary   Out of bounds exit 
		if (tx < 1 || tx > n || ty < 1 || ty > m)
			continue;
		// If it is currently O Just judge   If it is X Just skip 
		if (board[x][y] == 'O') {
			// If the next point is also O It shows that the two are connected   Directly flip the current point to X
			if (board[tx][ty] == 'O'&&book[tx][ty]==0) {
				book[tx][ty] = 1;
				board[x][y] = 'X';
				dfs(tx, ty);
			}
			// If the next point is X, That is, you need to judge whether the current point is surrounded by X, If so, turn it over , If there is O Then flip the current point 
			if (board[tx][ty] == 'X'&&book[tx][ty] == 0) {
				book[tx][ty] = 1;
				if (k ==2){
					board[x][y] ='X';
					return;
				}
				dfs(x, y);
				
			}
		}
	}
	return;
}
int main()
{
	int i, j;
	cin >> n >> m;
	// input data 
	for (i = 1;i <= n;i++)
		for (j = 1;j <= m;j++)
			cin >> board[i][j];
	// Except for the surrounding points 
	for(i=2;i<n-1;i++)
		for(j=2;j<m-1;j++)
			dfs(i, j);// Get into 
	cout << endl;
	// The output matrix 
	for (i = 1;i <= n;i++)
	{
		for (j = 1;j <= m;j++)
			cout << board[i][j] << " ";
		cout << endl;
	}
	return 0;
}