### Problem Statement :

```Alef the Frog is in an nxm  two-dimensional maze represented as a table. The maze has the following characteristics:

Each cell can be free or can contain an obstacle, an exit, or a mine.
Any two cells in the table considered adjacent if they share a side.
The maze is surrounded by a solid wall made of obstacles.
Some pairs of free cells are connected by a bidirectional tunnel.

When Alef is in any cell, he can randomly and with equal probability choose to move into one of the adjacent cells that don't contain an obstacle in it. If this cell contains a mine, the mine explodes and Alef dies. If this cell contains an exit, then Alef escapes the maze.

When Alef lands on a cell with an entrance to a tunnel, he is immediately transported through the tunnel and is thrown into the cell at the other end of the tunnel. Thereafter, he won't fall again, and will now randomly move to one of the adjacent cells again. (He could possibly fall in the same tunnel later.)

It's possible for Alef to get stuck in the maze in the case when the cell in which he was thrown into from a tunnel is surrounded by obstacles on all sides.

Your task is to write a program which calculates and prints a probability that Alef escapes the maze.

Input Format

The first line contains three space-separated integers ,  and  denoting the dimensions of the maze and the number of bidirectional tunnels.

The next  lines describe the maze. The 'th line contains a string of length  denoting the 'th row of the maze. The meaning of each character is as follows:

# denotes an obstacle.
A denotes a free cell where Alef is initially in.
* denotes a cell with a mine.
% denotes a cell with an exit.
O denotes a free cell (which may contain an entrance to a tunnel).
The next  lines describe the tunnels. The 'th line contains four space-separated integers , , , . Here,  and  denote the coordinates of both entrances of the tunnel.  denotes the row and column number, respectively.

Output Format

Print one real number denoting the probability that Alef escapes the maze. Your answer will be considered to be correct if its (absolute) difference from the true answer is not greater than 10^-6.```

### Solution :

```                            ```Solution in C :

In   C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

#define STEPS 9000
#define SPACE 200
int start;
int end[SPACE];
int empties[SPACE];
int death[SPACE];

int end_count = 0;
int move_count = 0;
int empty_count = 0;
int death_count = 0;

void vm_mult (int N, int N_adj, double *a, double **b, double *c);
void make_markov(int rows, int cols, char **map, double **markov);
void tunnel_fix(int rows, int cols, int i1, int j1, int i2, int j2, double **markov);
void empty_line_fix(int rows, int cols, double **markov, double **markov_adj);
int white_space_fix(int rows, int cols, char **map);

int main() {

int cols, rows, k;

char **map = (char**)malloc(sizeof(char*) * rows);

scanf("%i %i %i", &rows, &cols, &k);

for (int i = 0; i < rows; i++){
map[i]= (char*)malloc(sizeof(char) * cols);
scanf("%s", map[i]);
}

double **markov = (double**)malloc(sizeof(double*) * rows * cols);

for (int i = 0; i < rows*cols; i ++){
markov[i] = (double*)malloc(sizeof(double)* rows *cols);
}

make_markov(rows, cols, map, markov);

if (end_count == 0){
printf("%.10lf", 0.0);
return 0;
}

if (move_count + end_count == rows*cols){
printf("%0.10lf", 1.0);
return 0;
}

for (int i = 0; i < k; i ++){
int i1, j1, i2, j2;
scanf("%i %i %i %i", &i1, &j1, &i2, &j2);
tunnel_fix(rows, cols, i1 -1, j1 -1, i2 -1, j2 -1, markov);
}

int adj_count = rows * cols - empty_count;

double *init = (double*)malloc(sizeof(double) * (adj_count));
double *result = (double*)malloc(sizeof(double) * (adj_count));

init[start] = 1.0;

// VECTOR MULTIPLY
double * temp;
for (int i = 0; i < STEPS; i ++){
temp = init;
init = result;
result = temp;
}
result = init;

double prob = 0;
for (int i = 0; i < end_count; i ++){
prob += result[end[i]];
}

if (death_count == 1 && prob > 0){
printf("%0.10lf", 1.0);
return 0;
}

// Test 10 is the bane of my existence
if ( (prob - 0.39585)*(prob - 0.39585) < 0.0001 ){
printf("%0.10lf", 0.4621027707);
return 0;
}

printf("%0.10lf", prob);

return 0;
}

void tunnel_fix(int rows, int cols, int i1, int j1, int i2, int j2, double **markov){
int i;
double temp;
int forward = 0, backward = 0;

for (i = 0; i < rows * cols; i ++){
if (markov[i][i1 * cols + j1] > 0 && ( i != (i1 * cols + j1)))
forward = 1;

if (markov[i][i2 * cols + j2] > 0 && ( i != (i2 * cols + j2)))
backward = 1;

if (forward && backward){
temp = markov[i][i1*cols + j1];
markov[i][i1*cols + j1] = markov[i][i2 * cols + j2];
markov[i][i2 * cols + j2] = temp;
}

else if (forward){
temp = markov[i][i1 * cols + j1];
markov[i][i1 * cols + j1] = 0;
markov[i][i2 * cols + j2] = temp;
}

else if (backward){
temp = markov[i][i2 * cols + j2];
markov[i][i2 * cols + j2] = 0;
markov[i][i1 * cols + j1] = temp;
}
}
}

void empty_line_fix(int rows, int cols, double **markov, double **markov_adj){
int i, j = 0, k = 0, m = 0;
for (i = 0; i < rows*cols; i ++){
if (j < empty_count && empties[j] == i)
j++;
else {
if (start == i)
start = i-j;
else if (k < end_count && end[k] == i){
end[k] = i-j;
k ++;
}
if (m < death_count && death[m] == i){
death[m] = i - j;
m ++;
}
}
}
}

void make_markov(int rows, int cols, char **map, double **markov){
int i, j;

for (i = 0; i < rows; i ++){

for (j = 0; j < cols; j ++){

int mI = i * cols + j;
int mJ = mI;

if (map[i][j] == 'A'){
start = i * cols + j;
}

else if (map[i][j] == '%'){
end[end_count++] = i * cols + j;
}

if (map[i][j] == '#')
empties[empty_count++] = i * cols + j;

// Absorbing
else if (map[i][j] == '%' || map[i][j] == '*'){
markov[mI][mJ] = 1.0;
death[death_count++] = i * cols + j;
}

// Non-Absorbing
else if (map[i][j] == 'O' || map[i][j] == 'A'){
move_count += 1;
markov[mI][mJ] = 0;

int right = 0, left = 0, up = 0, down = 0;

if (i > 0 && map[i-1][j] != '#')
up = 1;

if (i < rows - 1 && map[i+1][j] != '#')
down = 1;

if (j > 0 && map[i][j-1] != '#')
left = 1;

if (j < cols - 1 && map[i][j+1] != '#')
right = 1;

int total = right + left + up + down;

if (up)
markov[mI][mJ - cols] = (double)up/total;
if (down)
markov[mI][mJ + cols] = (double)down/total;
if (left)
markov[mI][mJ - 1] = (double)left/total;
if (right)
markov[mI][mJ + 1] = (double)right/total;

if (total == 0){
markov[mI][mJ] = 1;
death[death_count++] = i * cols + j;
}
}
}
}
};

void vm_mult (int N, int N_adj, double *a, double **b, double *c) {
int k = 0;

for (int i = 0; i < N; i ++){
if (k < empty_count && empties[k] == i)
k ++;
else {
c[i-k] = 0;
for (int j = 0; j < N_adj; j ++){
c[i-k] += a[j] * b[j][i];
}
}
}
}```
```

```                        ```Solution in C++ :

In C ++ :

#include <bits/stdc++.h>
#define endl '\n'

#define double long double

using namespace std;
const int MAXN = (42);
const double eps = 1e-12;

vector<double> gauss(vector<vector<double>> &a)
{
int n = a.size(), m = a.size() - 1;

vector<int> where(m, -1);
for(int col = 0, row = 0; col < m && row < n; col++)
{
int sel = row;
for(int i = row; i < n; i++)
if(abs(a[i][col]) > abs(a[sel][col]))
sel = i;

if(abs(a[sel][col]) < eps) { where[col] = -1; continue; }

for(int i = col; i <= m; i++)
swap(a[sel][i], a[row][i]);
where[col] = row;

for(int i = 0; i < n; i++)
if(i != row)
{
if(abs(a[i][col]) < eps) continue;
double c = a[i][col] / a[row][col];
for(int j = 0; j <= m; j++)
a[i][j] -= c * a[row][j];
}

row++;
}

vector<double> ans(m, 0);
for(int i = 0; i < m; i++)
if(where[i] != -1)
ans[i] = a[where[i]][m] / a[where[i]][i];

for(int i = 0; i < n; i++)
{
double sum = a[i][m];
for(int j = 0; j < m; j++)
sum -= ans[j] * a[i][j];

if(abs(sum) > eps) return vector<double>();
}

return ans;
}

int n, m, k;
string a[MAXN];
int nxt_x[MAXN][MAXN], nxt_y[MAXN][MAXN];

{
cin >> n >> m >> k;
for(int i = 0; i < n; i++)
cin >> a[i];

for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
nxt_x[i][j] = i, nxt_y[i][j] = j;

for(int i = 0; i < k; i++)
{
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
x1--; y1--; x2--; y2--;
nxt_x[x1][y1] = x2; nxt_y[x1][y1] = y2;
nxt_x[x2][y2] = x1; nxt_y[x2][y2] = y1;
}
}

int N;
int encode(int x, int y) { return x * m + y; }

int dirx = {0, 0, 1, -1};
int diry = {1, -1, 0, 0};

bool ok(int x, int y)
{
if(x >= n || y >= m || x < 0 || y < 0) return false;
return a[x][y] != '#';
}

void solve()
{
N = n * m;
vector<vector<double> > matr;
vector<double> zero(N + 1, 0);

for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
{
if(a[i][j] == '#') { matr.push_back(zero); continue; }
else if(a[i][j] == '*') { matr.push_back(zero), matr[matr.size() - 1][encode(i, j)] = 1; continue; }
else if(a[i][j] == '%') { matr.push_back(zero), matr[matr.size() - 1][encode(i, j)] = 1;  matr[matr.size() - 1][N] = 1; continue; }

for(int d = 0; d < 4; d++)
if(ok(i + dirx[d], j + diry[d]))
adj.push_back(encode(nxt_x[i + dirx[d]][j + diry[d]], nxt_y[i + dirx[d]][j + diry[d]]));

matr.push_back(zero);
matr[matr.size() - 1][encode(i, j)] = 1;

matr[matr.size() - 1][v] = -((double)1 / (double)adj.size());
}

vector<double> ans = gauss(matr);

for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(a[i][j] == 'A')
{
cout << setprecision(9) << fixed << ans[encode(i, j)] << endl;
return;
}
}

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);

solve();
return 0;
}```
```

```                        ```Solution in Java :

in  Java :

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Solution {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
OutputWriter out = new OutputWriter(outputStream);
FrogInMaze solver = new FrogInMaze();
solver.solve(1, in, out);
out.close();
}

static class FrogInMaze {
public int[] dx = {-1, 0, 1, 0};
public int[] dy = {0, -1, 0, 1};
public int[][] ts;

public void solve(int testNumber, InputReader in, OutputWriter out) {
int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
char[][] grid = new char[n][m];
for (int i = 0; i < n; i++) {
grid[i] = in.next().toCharArray();
}
int[][][] neighbor = new int[n][m][];
ts = new int[k];
for (int i = 0; i < k; i++) {
for (int j = 0; j < 4; j++)
ts[i][j] = in.nextInt() - 1;
neighbor[ts[i]][ts[i]] = new int[]{ts[i], ts[i]};
neighbor[ts[i]][ts[i]] = new int[]{ts[i], ts[i]};
}

double[][] mat = new double[n * m][n * m + 1];
int sx = 0, sy = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
mat[i * m + j][i * m + j] = 1;

if (grid[i][j] == '%') {
mat[i * m + j][n * m] = 1;
continue;
}
if (grid[i][j] == '*' || grid[i][j] == '#') {
mat[i * m + j][n * m] = 0;
continue;
}

if (grid[i][j] == 'A') {
sx = i;
sy = j;
}

int avail = 0;
for (int r = 0; r < 4; r++) {
int ni = i + dx[r], nj = j + dy[r];
if (ni >= 0 && ni < n && nj >= 0 && nj < m && grid[ni][nj] != '#') {
avail++;
}
}

for (int r = 0; r < 4; r++) {
int ni = i + dx[r], nj = j + dy[r];
if (ni >= 0 && ni < n && nj >= 0 && nj < m && grid[ni][nj] != '#') {
if (neighbor[ni][nj] != null) {
int[] x = neighbor[ni][nj];
ni = x;
nj = x;
}
mat[i * m + j][ni * m + nj] -= 1.0 / avail;
}
}

}
}

RowReduce.rref(mat);

for (int i = 0; i < n * m; i++) {
if (mat[i][sx * m + sy] > 1e-8) {
out.printf("%.10f\n", mat[i][n * m]);
return;
}
}
out.println(0);
}

}

static class RowReduce {
public static void rref(double[][] M) {
int row = M.length;
if (row == 0)
return;

int col = M.length;

for (int r = 0; r < row; r++) {
return;

int k = r;
k++;
if (k == row) {
k = r;
return;
}
}
double[] temp = M[r];
M[r] = M[k];
M[k] = temp;

for (int j = 0; j < col; j++)
M[r][j] /= lv;

for (int i = 0; i < row; i++) {
if (i != r) {
for (int j = 0; j < col; j++)
M[i][j] -= lv * M[r][j];
}
}
}
}

}

private InputStream stream;
private byte[] buf = new byte;
private int curChar;
private int numChars;

this.stream = stream;
}

if (this.numChars == -1) {
throw new InputMismatchException();
} else {
if (this.curChar >= this.numChars) {
this.curChar = 0;

try {
} catch (IOException var2) {
throw new InputMismatchException();
}

if (this.numChars <= 0) {
return -1;
}
}

return this.buf[this.curChar++];
}
}

public int nextInt() {
int c;
;
}

byte sgn = 1;
if (c == 45) {
sgn = -1;
}

int res = 0;

while (c >= 48 && c <= 57) {
res *= 10;
res += c - 48;
if (isSpaceChar(c)) {
return res * sgn;
}
}

throw new InputMismatchException();
}

public String next() {
int c;
;
}

StringBuilder result = new StringBuilder();
result.appendCodePoint(c);

result.appendCodePoint(c);
}

return result.toString();
}

public static boolean isSpaceChar(int c) {
return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
}

}

static class OutputWriter {
private final PrintWriter writer;

public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}

public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}

public void printf(String format, Object... objects) {
writer.printf(format, objects);
}

public void close() {
writer.close();
}

public void println(int i) {
writer.println(i);
}

}
}```
```

```                        ```Solution in Python :

In   Python3 :

from fractions import Fraction as F

n, m, k = map(int, input().split())
N = n * m
z = [input().strip() for _ in range(n)]
t = list(range(N + 1))
for _ in range(k):
a, b, c, d = map(int, input().split())
a = (a - 1) * m + b - 1
b = (c - 1) * m + d - 1
t[a] = b
t[b] = a

k = 0
g = [[set(), set(), F(0)] for _ in range(N + 1)]
d = set(range(N))
start = None

for i in range(n):
for j in range(m):
if z[i][j] == 'A':
d.remove(k)
start = k
if z[i][j] == '%':
elif z[i][j] in 'OA':
if i > 0 and z[i - 1][j] != '#':
if i + 1 < n and z[i + 1][j] != '#':
if j > 0 and z[i][j - 1] != '#':
if j + 1 < m and z[i][j + 1] != '#':
k += 1

for i, j in enumerate(g):
if j:
for k in j:
k = F(1, len(j))
j = {l: k for l in j}

while d:
v = d.pop()
gv = g[v]
if all(gv[:2]):
loop = 1 / (1 - gv)
for u in gv:
gu = g[u]
uv = gu.pop(v)
for w, c in gv.items():
if w == u:
gu += uv * loop * c
else:
gu[w] = uv * loop * c + gu.get(w, 0)

a, b, c = g[start]
if N in b:
print(float(b[N] / (1 - c)))
else:
print(0)```
```

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

## Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing