给定两个整数R和C ，分别表示矩阵中的行和列数，以及两个整数X和Y ，任务是找到从给定单元格到矩阵的所有其他单元格的最小距离。
例子：
Input: R = 5, C = 5, X = 2, Y = 2
Output:
2 2 2 2 2
2 1 1 1 2
2 1 0 1 2
2 1 1 1 2
2 2 2 2 2
Input: R = 5, C = 5, X = 1, Y = 1
Output:
1 1 1 2 3
1 0 1 2 3
1 1 1 2 3
2 2 2 2 3
3 3 3 3 3
方法：
请按照以下步骤解决问题：
- 将初始像元的距离指定为0 。
- 初始化一个队列，并将对{X，Y}插入到Queue中。
- 迭代直到队列不为空，并使用BFS技术为每个未访问的单元分配当前距离并将索引{i，j}插入队列。
- 在末尾打印每个单元格的距离。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
int mat[1001][1001];
int r, c, x, y;
// Stores the accessible directions
int dx[] = { 0, -1, -1, -1, 0, 1, 1, 1 };
int dy[] = { 1, 1, 0, -1, -1, -1, 0, 1 };
// Function to find the minimum distance from a
// given cell to all other cells in the matrix
void FindMinimumDistance()
{
// Stores the accessible cells
// from current cell
queue > q;
// Insert pair (x, y)
q.push({ x, y });
mat[x][y] = 0;
// Iterate untill queue is empty
while (!q.empty()) {
// Extract the pair
x = q.front().first;
y = q.front().second;
// Pop them
q.pop();
for (int i = 0; i < 8; i++) {
int a = x + dx[i];
int b = y + dy[i];
// Checking boundary condition
if (a < 0 || a >= r || b >= c || b < 0)
continue;
// If the cell is not visited
if (mat[a][b] == 0) {
// Assign the minimum distance
mat[a][b] = mat[x][y] + 1;
// Insert the traversed neighbour
// into the queue
q.push({ a, b });
}
}
}
}
// Driver Code
int main()
{
r = 5, c = 5, x = 1, y = 1;
int t = x;
int l = y;
mat[x][y] = 0;
FindMinimumDistance();
mat[t][l] = 0;
// Print the required distances
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
cout << mat[i][j] << " ";
}
cout << endl;
}
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
static int [][]mat = new int[1001][1001];
static int r, c, x, y;
// Stores the accessible directions
static int dx[] = { 0, -1, -1, -1, 0, 1, 1, 1 };
static int dy[] = { 1, 1, 0, -1, -1, -1, 0, 1 };
// Function to find the minimum distance from a
// given cell to all other cells in the matrix
static void FindMinimumDistance()
{
// Stores the accessible cells
// from current cell
Queue q = new LinkedList<>();
// Insert pair (x, y)
q.add(new pair(x, y));
mat[x][y] = 0;
// Iterate untill queue is empty
while (!q.isEmpty())
{
// Extract the pair
x = q.peek().first;
y = q.peek().second;
// Pop them
q.remove();
for(int i = 0; i < 8; i++)
{
int a = x + dx[i];
int b = y + dy[i];
// Checking boundary condition
if (a < 0 || a >= r ||
b >= c || b < 0)
continue;
// If the cell is not visited
if (mat[a][b] == 0)
{
// Assign the minimum distance
mat[a][b] = mat[x][y] + 1;
// Insert the traversed neighbour
// into the queue
q.add(new pair(a, b));
}
}
}
}
// Driver Code
public static void main(String[] args)
{
r = 5; c = 5; x = 1; y = 1;
int t = x;
int l = y;
mat[x][y] = 0;
FindMinimumDistance();
mat[t][l] = 0;
// Print the required distances
for(int i = 0; i < r; i++)
{
for(int j = 0; j < c; j++)
{
System.out.print(mat[i][j] + " ");
}
System.out.println();
}
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 program to implement
# the above approach
mat = [[0 for x in range(1001)]
for y in range(1001)]
# Stores the accessible directions
dx = [ 0, -1, -1, -1, 0, 1, 1, 1 ]
dy = [ 1, 1, 0, -1, -1, -1, 0, 1 ]
# Function to find the minimum distance
# from a given cell to all other cells
# in the matrix
def FindMinimumDistance():
global x, y, r, c
# Stores the accessible cells
# from current cell
q = []
# Insert pair (x, y)
q.append([x, y])
mat[x][y] = 0
# Iterate untill queue is empty
while(len(q) != 0):
# Extract the pair
x = q[0][0]
y = q[0][1]
# Pop them
q.pop(0)
for i in range(8):
a = x + dx[i]
b = y + dy[i]
# Checking boundary condition
if(a < 0 or a >= r or
b >= c or b < 0):
continue
# If the cell is not visited
if(mat[a][b] == 0):
# Assign the minimum distance
mat[a][b] = mat[x][y] + 1
# Insert the traversed neighbour
# into the queue
q.append([a, b])
# Driver Code
r = 5
c = 5
x = 1
y = 1
t = x
l = y
mat[x][y] = 0
FindMinimumDistance()
mat[t][l] = 0
# Print the required distances
for i in range(r):
for j in range(c):
print(mat[i][j], end = " ")
print()
# This code is contributed by Shivam SinghC#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
static int [,]mat = new int[1001, 1001];
static int r, c, x, y;
// Stores the accessible directions
static int []dx = { 0, -1, -1, -1, 0, 1, 1, 1 };
static int []dy = { 1, 1, 0, -1, -1, -1, 0, 1 };
// Function to find the minimum distance from a
// given cell to all other cells in the matrix
static void FindMinimumDistance()
{
// Stores the accessible cells
// from current cell
Queue q = new Queue();
// Insert pair (x, y)
q.Enqueue(new pair(x, y));
mat[x, y] = 0;
// Iterate untill queue is empty
while (q.Count != 0)
{
// Extract the pair
x = q.Peek().first;
y = q.Peek().second;
// Pop them
q.Dequeue();
for(int i = 0; i < 8; i++)
{
int a = x + dx[i];
int b = y + dy[i];
// Checking boundary condition
if (a < 0 || a >= r ||
b >= c || b < 0)
continue;
// If the cell is not visited
if (mat[a, b] == 0)
{
// Assign the minimum distance
mat[a, b] = mat[x, y] + 1;
// Insert the traversed neighbour
// into the queue
q.Enqueue(new pair(a, b));
}
}
}
}
// Driver Code
public static void Main(String[] args)
{
r = 5; c = 5; x = 1; y = 1;
int t = x;
int l = y;
mat[x, y] = 0;
FindMinimumDistance();
mat[t, l] = 0;
// Print the required distances
for(int i = 0; i < r; i++)
{
for(int j = 0; j < c; j++)
{
Console.Write(mat[i, j] + " ");
}
Console.WriteLine();
}
}
}
// This code is contributed by shikhasingrajput 输出：
1 1 1 2 3
1 0 1 2 3
1 1 1 2 3
2 2 2 2 3
3 3 3 3 3
时间复杂度： O(R * C)
辅助空间： O(R * C)