Pacific Atlantic current problem

There is one with The Pacific Ocean and The Atlantic ocean m x n Bordering rectangular islands . The Pacific Ocean Touch the left and upper edges of the island , The Atlantic ocean Touch the right and lower edges of the island .

The island is divided into square cells . Given a m x n Integer matrix heights, among heights[r][c] Represents the cell at the coordinates Altitude (r, c).

There is a lot of rainfall on the island , If the height of adjacent cells is less than or equal to The height of the current cell , The rain can go straight north 、 south 、 In the east 、 West flow to adjacent cells . Water can flow into the ocean from any cell adjacent to the ocean .

return Grid coordinates Of 2D list result, among Indicates that rainwater can flow from the cell The Pacific Ocean and The Atlantic ocean .result[i] = [ri, ci](ri, ci)

 

Example 1：

 Input ： Height  = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4, 5],[5,1,1,2,4]]  Output ： [[0,4],[1,3],[1,4],[2,2],[3,0],[3 ,1],[4,0]] 

Example 2：

 Input ： Height  = [[2,1],[1,2]]  Output ： [[0,0],[0,1],[1,0],[1,1]] 

  Error code

#include<iostream>
using namespace std;
int height[201][201], book[201][201];
int m, n,num=1;
// Current point coordinates 
void dfs(int x, int y) {
	int i, j, k, tx, ty;

	// Define four directions 
	int next[4][2] = {
		{0,1},{1,0},{0,-1},{-1,0}
	};// Right down left up 

	// Try whether you can walk in four directions 
	for (k = 0;k < 4;k++) {
		tx = x + next[k][0];
		ty = y + next[k][1];
	
		// If the right point is larger than the current point, it can flow through and then judge whether the next point meets the conditions , Find the right book All set 
		if (tx>=0 && ty>=0 && tx<n  && ty<m  &&height[tx][ty] >= height[x][y]&&book[tx][ty]==0) {
			book[tx][ty] =1;
			dfs(tx, ty);
		}
	}
	return;
}
int main() {
	int i, j, a[201][201] = { 0 }, b[201][201];
	// Enter row and column 
	cin >> n >> m;
	// input data 
	for (i = 0;i < n;i++)
		for (j = 0;j < m;j++)
			cin >> height[i][j];
	// Points that meet the conditions enter the search （ The surrounding points are divided into four parts ）
	// On 
	
	for (i = 0;i < n;i++) {
		book[0][i] = 1;
        dfs(0, i);
	}
		
	// Left 
	
	for (i = 1;i < n;i++){
		book[i][0] = 1;
		dfs(i, 0);}

	for (i = 0;i < n;i++)
		for (j = 0;j < m;j++)
			a[i][j] = book[i][j];// take book All passed in are assigned to a Array , The staging 



     // Empty book Tag array 
	
	for (i = 0;i < n;i++) {
		for (j = 0;j < m;j++)
			book[i][j] = { 0 };
	}

	cout << endl;
	// Next 
	for (i = 1;i < n;i++) {
		book[4][i] = 1;
		dfs(n-1, i);
	}
		
	// Right 
	
	for (i = 1;i < n - 1;i++) {
		book[i][4] = 1;
		dfs(i, n-1);
	}
		

	// look for a and book Output the overlapped part of the array 
	for (i = 0;i < n;i++)
		for (j = 0;j < m;j++)
		{
			if (a[i][j] == book[i][j])
				cout << i << " " << j << endl;
		}

	return 0;
}



5 5
1 2 2 3 5
3 2 3 4 4
2 4 5 3 1
6 7 1 4 5
5 1 1 2 4