在银行中找到与守卫的最短距离
给定一个由“O”、“G”和“W”填充的矩阵，其中“O”代表开放空间，“G”代表守卫，“W”代表银行的墙壁。将矩阵中的所有 O 替换为与守卫的最短距离，而不能穿过任何墙壁。此外，在输出矩阵中用 0 替换守卫，用 -1 替换墙壁。
M x N 矩阵的预期时间复杂度为 O(MN)。
例子：
O ==> Open Space
G ==> Guard
W ==> Wall
Input:
O O O O G
O W W O O
O O O W O
G W W W O
O O O O G
Output:
3 3 2 1 0
2 -1 -1 2 1
1 2 3 -1 2
0 -1 -1 -1 1
1 2 2 1 0这个想法是做BFS。我们首先将所有包含警卫的单元排入队列并循环直到队列不为空。对于循环的每次迭代，我们将前面的单元格从队列中取出，对于它的四个相邻单元格中的每一个，如果单元格是一个开放区域并且尚未计算它与守卫的距离，我们会更新它的距离并将其入队。最后在 BFS 过程结束后，我们打印距离矩阵。
以下是上述想法的实现 -
C++
// C++ program to replace all of the O's in the matrix
// with their shortest distance from a guard
#include
using namespace std;
// store dimensions of the matrix
#define M 5
#define N 5
// An Data Structure for queue used in BFS
struct queueNode
{
// i, j and distance stores x and y-coordinates
// of a matrix cell and its distance from guard
// respectively
int i, j, distance;
};
// These arrays are used to get row and column
// numbers of 4 neighbors of a given cell
int row[] = { -1, 0, 1, 0};
int col[] = { 0, 1, 0, -1 };
// return true if row number and column number
// is in range
bool isValid(int i, int j)
{
if ((i < 0 || i > M - 1) || (j < 0 || j > N - 1))
return false;
return true;
}
// return true if current cell is an open area and its
// distance from guard is not calculated yet
bool isSafe(int i, int j, char matrix[][N], int output[][N])
{
if (matrix[i][j] != 'O' || output[i][j] != -1)
return false;
return true;
}
// Function to replace all of the O's in the matrix
// with their shortest distance from a guard
void findDistance(char matrix[][N])
{
int output[M][N];
queue q;
// finding Guards location and adding into queue
for (int i = 0; i < M; i++)
{
for (int j = 0; j < N; j++)
{
// initialize each cell as -1
output[i][j] = -1;
if (matrix[i][j] == 'G')
{
queueNode pos = {i, j, 0};
q.push(pos);
// guard has 0 distance
output[i][j] = 0;
}
}
}
// do till queue is empty
while (!q.empty())
{
// get the front cell in the queue and update
// its adjacent cells
queueNode curr = q.front();
int x = curr.i, y = curr.j, dist = curr.distance;
// do for each adjacent cell
for (int i = 0; i < 4; i++)
{
// if adjacent cell is valid, has path and
// not visited yet, en-queue it.
if (isSafe(x + row[i], y + col[i], matrix, output)
&& isValid(x + row[i], y + col[i]))
{
output[x + row[i]][y + col[i]] = dist + 1;
queueNode pos = {x + row[i], y + col[i], dist + 1};
q.push(pos);
}
}
// dequeue the front cell as its distance is found
q.pop();
}
// print output matrix
for (int i = 0; i < M; i++)
{
for (int j = 0; j < N; j++)
cout << std::setw(3) << output[i][j];
cout << endl;
}
}
// Driver code
int main()
{
char matrix[][N] =
{
{'O', 'O', 'O', 'O', 'G'},
{'O', 'W', 'W', 'O', 'O'},
{'O', 'O', 'O', 'W', 'O'},
{'G', 'W', 'W', 'W', 'O'},
{'O', 'O', 'O', 'O', 'G'}
};
findDistance(matrix);
return 0;
} Java
// Java program to replace all of the O's
// in the matrix with their shortest
// distance from a guard
package Graphs;
import java.util.LinkedList;
import java.util.Queue;
public class MinDistanceFromaGuardInBank{
// Store dimensions of the matrix
int M = 5;
int N = 5;
class Node
{
int i, j, dist;
Node(int i, int j, int dist)
{
this.i = i;
this.j = j;
this.dist = dist;
}
}
// These arrays are used to get row
// and column numbers of 4 neighbors
// of a given cell
int row[] = { -1, 0, 1, 0 };
int col[] = { 0, 1, 0, -1 };
// Return true if row number and
// column number is in range
boolean isValid(int i, int j)
{
if ((i < 0 || i > M - 1) ||
(j < 0 || j > N - 1))
return false;
return true;
}
// Return true if current cell is
// an open area and its distance
// from guard is not calculated yet
boolean isSafe(int i, int j, char matrix[][],
int output[][])
{
if (matrix[i][j] != 'O' ||
output[i][j] != -1)
return false;
return true;
}
// Function to replace all of the O's
// in the matrix with their shortest
// distance from a guard
void findDistance(char matrix[][])
{
int output[][] = new int[M][N];
Queue q = new LinkedList();
// Finding Guards location and
// adding into queue
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
// Initialize each cell as -1
output[i][j] = -1;
if (matrix[i][j] == 'G')
{
q.add(new Node(i, j, 0));
// Guard has 0 distance
output[i][j] = 0;
}
}
}
// Do till queue is empty
while (!q.isEmpty())
{
// Get the front cell in the queue
// and update its adjacent cells
Node curr = q.peek();
int x = curr.i;
int y = curr.j;
int dist = curr.dist;
// Do for each adjacent cell
for (int i = 0; i < 4; i++)
{
// If adjacent cell is valid, has
// path and not visited yet,
// en-queue it.
if (isValid(x + row[i], y + col[i]))
{
if (isSafe(x + row[i], y + col[i],
matrix, output))
{
output[x + row[i]][y + col[i]] = dist + 1;
q.add(new Node(x + row[i],
y + col[i],
dist + 1));
}
}
}
// Dequeue the front cell as
// its distance is found
q.poll();
}
// Print output matrix
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
System.out.print(output[i][j] + " ");
}
System.out.println();
}
}
// Driver code
public static void main(String args[])
{
char matrix[][] = { { 'O', 'O', 'O', 'O', 'G' },
{ 'O', 'W', 'W', 'O', 'O' },
{ 'O', 'O', 'O', 'W', 'O' },
{ 'G', 'W', 'W', 'W', 'O' },
{ 'O', 'O', 'O', 'O', 'G' } };
MinDistanceFromaGuardInBank g =
new MinDistanceFromaGuardInBank();
g.findDistance(matrix);
}
}
// This code is contributed by Shobhit Yadav Python3
# Python3 program to replace all of the O's in the matrix
# with their shortest distance from a guard
from collections import deque as queue
# store dimensions of the matrix
M = 5
N = 5
# These arrays are used to get row and column
# numbers of 4 neighbors of a given cell
row = [-1, 0, 1, 0]
col = [0, 1, 0, -1]
# return true if row number and column number
# is in range
def isValid(i, j):
if ((i < 0 or i > M - 1) or (j < 0 or j > N - 1)):
return False
return True
# return true if current cell is an open area and its
# distance from guard is not calculated yet
def isSafe(i, j,matrix, output):
if (matrix[i][j] != 'O' or output[i][j] != -1):
return False
return True
# Function to replace all of the O's in the matrix
# with their shortest distance from a guard
def findDistance(matrix):
output = [[ -1 for i in range(N)]for i in range(M)]
q = queue()
# finding Guards location and adding into queue
for i in range(M):
for j in range(N):
# initialize each cell as -1
output[i][j] = -1
if (matrix[i][j] == 'G'):
pos = [i, j, 0]
q.appendleft(pos)
# guard has 0 distance
output[i][j] = 0
# do till queue is empty
while (len(q) > 0):
# get the front cell in the queue and update
# its adjacent cells
curr = q.pop()
x, y, dist = curr[0], curr[1], curr[2]
# do for each adjacent cell
for i in range(4):
# if adjacent cell is valid, has path and
# not visited yet, en-queue it.
if isValid(x + row[i], y + col[i]) and isSafe(x + row[i], y + col[i], matrix, output) :
output[x + row[i]][y + col[i]] = dist + 1
pos = [x + row[i], y + col[i], dist + 1]
q.appendleft(pos)
# print output matrix
for i in range(M):
for j in range(N):
if output[i][j] > 0:
print(output[i][j], end=" ")
else:
print(output[i][j],end=" ")
print()
# Driver code
matrix = [['O', 'O', 'O', 'O', 'G'],
['O', 'W', 'W', 'O', 'O'],
['O', 'O', 'O', 'W', 'O'],
['G', 'W', 'W', 'W', 'O'],
['O', 'O', 'O', 'O', 'G']]
findDistance(matrix)
# This code is contributed by mohit kumar 29C#
// C# program to replace all of the O's
// in the matrix with their shortest
// distance from a guard
using System;
using System.Collections.Generic;
public class Node
{
public int i, j, dist;
public Node(int i, int j, int dist)
{
this.i = i;
this.j = j;
this.dist = dist;
}
}
public class MinDistanceFromaGuardInBank
{
// Store dimensions of the matrix
static int M = 5;
static int N = 5;
// These arrays are used to get row
// and column numbers of 4 neighbors
// of a given cell
static int[] row = { -1, 0, 1, 0 };
static int[] col = { 0, 1, 0, -1 };
// Return true if row number and
// column number is in range
static bool isValid(int i, int j)
{
if ((i < 0 || i > M - 1) || (j < 0 || j > N - 1))
return false;
return true;
}
// Return true if current cell is
// an open area and its distance
// from guard is not calculated yet
static bool isSafe(int i, int j, char[,] matrix,int[,] output)
{
if (matrix[i,j] != 'O' || output[i,j] != -1)
{
return false;
}
return true;
}
// Function to replace all of the O's
// in the matrix with their shortest
// distance from a guard
static void findDistance(char[,] matrix)
{
int[,] output = new int[M,N];
Queue q = new Queue();
// Finding Guards location and
// adding into queue
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
// Initialize each cell as -1
output[i, j] = -1;
if (matrix[i, j] == 'G')
{
q.Enqueue(new Node(i, j, 0));
// Guard has 0 distance
output[i, j] = 0;
}
}
}
// Do till queue is empty
while (q.Count != 0)
{
// Get the front cell in the queue
// and update its adjacent cells
Node curr = q.Peek();
int x = curr.i;
int y = curr.j;
int dist = curr.dist;
// Do for each adjacent cell
for (int i = 0; i < 4; i++)
{
// If adjacent cell is valid, has
// path and not visited yet,
// en-queue it.
if (isValid(x + row[i], y + col[i]))
{
if (isSafe(x + row[i], y + col[i],matrix, output))
{
output[x + row[i] , y + col[i]] = dist + 1;
q.Enqueue(new Node(x + row[i],y + col[i],dist + 1));
}
}
}
// Dequeue the front cell as
// its distance is found
q.Dequeue();
}
// Print output matrix
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
Console.Write(output[i,j] + " ");
}
Console.WriteLine();
}
}
// Driver code
static public void Main ()
{
char[,] matrix ={ { 'O', 'O', 'O', 'O', 'G' },
{ 'O', 'W', 'W', 'O', 'O' },
{ 'O', 'O', 'O', 'W', 'O' },
{ 'G', 'W', 'W', 'W', 'O' },
{ 'O', 'O', 'O', 'O', 'G' } };
findDistance(matrix);
}
}
// This code is contributed by avanitrachhadiya2155 Javascript
输出：
3 3 2 1 0
2 -1 -1 2 1
1 2 3 -1 2
0 -1 -1 -1 1
1 2 2 1 0