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# Answer to Question #44943 in C++ for Rahul Jobanputra

Question #44943
Given a square maze (A) of dimension N, every entry (Aij) in the maze is either an open cell 'O' or a wall 'X'. A rat can travel to its adjacent locations (left, right, top and bottom), but to reach a cell, it must be open. Given the locations of R rats, can you find out whether all the rats can reach others or not?
Input Format:

Input will consist of three parts, viz.

1. Size of the maze (N)
2. The maze itself (A = N * N)
3. Number of rats (R)
4. Location of R rats (Xi, Yi)

Note:
(Xi,Yi) will represents the location of the i-th rat.
Locations are 1-index based.

Output Format:

Print &quot;Yes&quot; if the rats can reach each other, else print &quot;No&quot;

Constraints:
1&lt;=N&lt;=350

Aij = {'O','X'}

1&lt;=R&lt;=N*N

1&lt;=Xi&lt;=N

1&lt;=Yi&lt;=N
1
2014-08-21T03:36:39-0400
//Answer on Question#44943 - Progamming - C++
#include <iostream>
#include <string>
using namespace std;
const int MAX_N = 350;
// depth-first-search algorithm
// takes the maze, the size of the maze. the current position in the maze
int dfs(string maze[], int N, int x, int y) {
int rats = 0;
if (x < 0 || x >= N) return rats; // the position is not in the maze
if (y < 0 || y >= N) return rats; // the position is not in the maze
if (maze[x][y] != 'O' && maze[x][y] != 'R') return rats; // the position isn't open
if (maze[x][y] == 'R') ++rats;
maze[x][y] = 'X'; // make the cell closed beacuse we already visited it
static const int dx[4] = {-1, 0, 1, 0};
static const int dy[4] = {0, 1, 0, -1};
for (int i = 0; i < 4; ++i) {
int nx = x + dx[i];
int ny = y + dy[i];
rats += dfs(maze, N, nx, ny);
}
return rats;
}
int main() {
int N;
cin >> N;
string maze[MAX_N];
for (int i = 0; i < N; ++i) {
cin >> maze[i];
}
int R;
cin >> R;
for (int i = 0; i < R; ++i) {
int X, Y;
cin >> X >> Y;
--X; --Y;
maze[X][Y] = 'R';
}
// run the depth-first-search algorithm for unvisited cell
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
int rats = dfs(maze, N, i, j);
if (rats == R) { // all rats are in the same connected component
cout << "Yes" << endl;
return 0;
} else if (rats > 0) { // there is a rat in a component with not all other rats
cout << "No" << endl;
return 0;
}
}
}
}

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Assignment Expert
03.08.15, 17:35

Dear rethna, You're welcome. We are glad to be helpful. If you liked our service please press like-button beside answer field. Thank you!

rethna
03.08.15, 12:06

than you sir