给定一个由0和1组成的大小为N的矩阵，任务是找到用1填充整个矩阵所需的最短时间。矩阵中每一个瞬间的1 ，可以将其八个相邻单元格中的所有0转换为1 ，即(i,j)处的1可以将所有0转换为1在位置(i, j-1) , (i, j+1) , (i-1, j) , (i+1, j) , (i-1, j-1) , (i-1, j+1) , (i+1, j-1) , (i+1, j+1) 。
例子：
Input: N = 3, mtrx[][] = {{1,0,0},{0,0,1},{0,0,0}}
Output: 2
Explanation:
Initially the matrix appears to be
1, 0, 0
0, 0, 1
0, 0, 0
After the first instant of time, the new matrix is
1, 1, 1
1, 1, 1
0, 1, 1
After the 2nd instant the remaining 0 is converted to 1.
Input: N = 5,
mtrx[][] = {{0,0,0,0,0},
{1,0,0,0,0},
{0,0,0,0,0},
{0,0,1,0,0},
{0,0,0,1,0}}
Output: 3
方法：
为了解决这个问题，我们使用了 BFS 方法。遍历矩阵并将矩阵的索引以 1 初始存储在队列中。循环直到队列为空并将所有队列元素的相邻有效单元格转换为 1。导致至少一个 0 到 1 转换的 BFS 遍历的级别数就是答案。
下面的代码是上述方法的实现：
C++
// C++ program to calculate the number of steps
// in fill all entire matrix with '1'
#include
using namespace std;
// Function to return total number
// of steps required to fill
// entire matrix with '1'
int numberOfSteps(int n,vector> mtrx)
{
// Queue to store indices with 1
queue> q;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (mtrx[i][j] == 1) {
q.push({i,j});
}
}
}
// Intialise step variable as zero
int step = 0 ;
// BFS approach
while (!q.empty())
{
int qsize = q.size();
// Visit all indices with 1 and
// fill its neighbouring cells by 1
while (qsize--)
{
pair p = q.front();
q.pop();
int i = p.first;
int j = p.second;
// Convert the neighbour from '0' to '1'
if((j > 0) && mtrx[i][j-1] == 0)
{
mtrx[i][j-1] = 1;
q.push({i,j-1});
}
// Convert the neighbour from '0' to '1'
if((i < n-1) && mtrx[i+1][j] == 0)
{
mtrx[i+1][j] = 1;
q.push({i+1,j});
}
// Convert the neighbour from '0' to '1'
if((j < n-1) && mtrx[i][j+1] == 0)
{
mtrx[i][j+1] = 1;
q.push({i,j + 1});
}
// Convert the neighbour from '0' to '1'
if((i > 0) && mtrx[i-1][j] == 0)
{
mtrx[i-1][j] = 1;
q.push({i-1,j});
}
// Convert the neighbour from '0' to '1'
if((i > 0) && (j > 0) &&
mtrx[i-1][j-1] == 0)
{
mtrx[i-1][j-1] = 1;
q.push({i-1,j-1});
}
// Convert the neighbour from '0' to '1'
if((i > 0) && (j < (n-1)) &&
mtrx[i-1][j+1] == 0)
{
mtrx[i-1][j+1] = 1;
q.push({i-1,j+1});
}
// Convert the neighbour from '0' to '1'
if((i < (n-1)) && (j < (n-1)) &&
mtrx[i+1][j+1] == 0)
{
mtrx[i+1][j+1] = 1;
q.push({i+1,j + 1});
}
// Convert the neighbour from '0' to '1'
if((i < (n-1)) && (j > 0) &&
mtrx[i+1][j-1] == 0)
{
mtrx[i+1][j-1] = 1;
q.push({i+1,j-1});
}
}
//count steps
step++;
}
return step-1;
}
// Driver code
int main()
{
int n = 5 ; //dimension of matrix NXN
vector> mtrx = {{0,0,0,0,0},
{1,0,0,0,0},
{0,0,0,0,0},
{0,0,1,0,0},
{0,0,0,1,0}};
// Print number of steps
cout << numberOfSteps(n, mtrx);
return 0;
} Java
// Java program to calculate the
// number of steps in fill all
// entire matrix with '1'
import java.util.*;
class GFG{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to return total number
// of steps required to fill
// entire matrix with '1'
static int numberOfSteps(int n, int[][] mtrx)
{
// Queue to store indices with 1
Queue q = new LinkedList<>();
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
if (mtrx[i][j] == 1)
{
q.add(new pair(i, j));
}
}
}
// Intialise step variable as zero
int step = 0;
// BFS approach
while (!q.isEmpty())
{
int qsize = q.size();
// Visit all indices with 1 and
// fill its neighbouring cells by 1
while (qsize-- > 0)
{
pair p = q.peek();
q.remove();
int i = p.first;
int j = p.second;
// Convert the neighbour from '0' to '1'
if ((j > 0) && mtrx[i][j - 1] == 0)
{
mtrx[i][j - 1] = 1;
q.add(new pair(i, j - 1));
}
// Convert the neighbour from '0' to '1'
if ((i < n - 1) && mtrx[i + 1][j] == 0)
{
mtrx[i + 1][j] = 1;
q.add(new pair(i + 1, j));
}
// Convert the neighbour from '0' to '1'
if ((j < n - 1) && mtrx[i][j + 1] == 0)
{
mtrx[i][j + 1] = 1;
q.add(new pair(i, j + 1));
}
// Convert the neighbour from '0' to '1'
if ((i > 0) && mtrx[i - 1][j] == 0)
{
mtrx[i - 1][j] = 1;
q.add(new pair(i - 1, j));
}
// Convert the neighbour from '0' to '1'
if ((i > 0) && (j > 0) &&
mtrx[i - 1][j - 1] == 0)
{
mtrx[i - 1][j - 1] = 1;
q.add(new pair(i - 1, j - 1));
}
// Convert the neighbour from '0' to '1'
if ((i > 0) && (j < (n - 1)) &&
mtrx[i - 1][j + 1] == 0)
{
mtrx[i - 1][j + 1] = 1;
q.add(new pair(i - 1, j + 1));
}
// Convert the neighbour from '0' to '1'
if ((i < (n - 1)) && (j < (n - 1)) &&
mtrx[i + 1][j + 1] == 0)
{
mtrx[i + 1][j + 1] = 1;
q.add(new pair(i + 1, j + 1));
}
// Convert the neighbour from '0' to '1'
if ((i < (n - 1)) && (j > 0) &&
mtrx[i + 1][j - 1] == 0)
{
mtrx[i + 1][j - 1] = 1;
q.add(new pair(i + 1, j - 1));
}
}
// Count steps
step++;
}
return step - 1;
}
// Driver code
public static void main(String[] args)
{
// Dimension of matrix NXN
int n = 5;
int [][]mtrx = { { 0, 0, 0, 0, 0 },
{ 1, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 1, 0, 0 },
{ 0, 0, 0, 1, 0 } };
// Print number of steps
System.out.print(numberOfSteps(n, mtrx));
}
}
// This code is contributed by Amit Katiyar Python 3
# Python 3 program to calculate the number of steps
# in fill all entire matrix with '1'
# Function to return total number
# of steps required to fill
# entire matrix with '1'
def numberOfSteps(n,mtrx):
# Queue to store indices with 1
q = []
for i in range(n):
for j in range(n):
if (mtrx[i][j] == 1):
q.append([i,j])
# Intialise step variable as zero
step = 0
# BFS approach
while (len(q)):
qsize = len(q)
# Visit all indices with 1 and
# fill its neighbouring cells by 1
while(qsize):
p = q[0]
q.remove(q[0])
i = p[0]
j = p[1]
# Convert the neighbour from '0' to '1'
if((j > 0) and mtrx[i][j - 1] == 0):
mtrx[i][j - 1] = 1
q.append([i, j - 1])
# Convert the neighbour from '0' to '1'
if((i < n-1) and mtrx[i + 1][j] == 0):
mtrx[i + 1][j] = 1
q.append([i + 1, j])
# Convert the neighbour from '0' to '1'
if((j < n-1) and mtrx[i][j + 1] == 0):
mtrx[i][j + 1] = 1
q.append([i, j + 1])
# Convert the neighbour from '0' to '1'
if((i > 0) and mtrx[i - 1][j] == 0):
mtrx[i - 1][j] = 1
q.append([i - 1, j])
# Convert the neighbour from '0' to '1'
if((i > 0) and (j > 0) and
mtrx[i - 1][j - 1] == 0):
mtrx[i - 1][j - 1] = 1
q.append([i - 1, j - 1])
# Convert the neighbour from '0' to '1'
if((i > 0) and (j < (n-1)) and
mtrx[i - 1][j + 1] == 0):
mtrx[i - 1][j + 1] = 1
q.append([i - 1, j + 1])
# Convert the neighbour from '0' to '1'
if((i < (n - 1)) and (j < (n - 1)) and
mtrx[i + 1][j + 1] == 0):
mtrx[i + 1][j + 1] = 1
q.append([i + 1, j + 1])
# Convert the neighbour from '0' to '1'
if((i < (n - 1)) and (j > 0) and
mtrx[i + 1][j - 1] == 0):
mtrx[i + 1][j - 1] = 1
q.append([i + 1,j - 1])
qsize -= 1
#count steps
step += 1
return step-1
# Driver code
if __name__ == '__main__':
#dimension of matrix NXN
n = 5
mtrx = [[ 0, 0, 0, 0, 0 ],
[ 1, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0 ],
[ 0, 0, 1, 0, 0 ],
[ 0, 0, 0, 1, 0 ]]
# Print number of steps
print(numberOfSteps(n, mtrx))
# This code is contributed by SamarthC#
// C# program to calculate the
// number of steps in fill all
// entire matrix with '1'
using System;
using System.Collections.Generic;
class GFG{
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to return total number
// of steps required to fill
// entire matrix with '1'
static int numberOfSteps(int n,
int[,] mtrx)
{
// Queue to store indices with 1
Queue q = new Queue();
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
if (mtrx[i, j] == 1)
{
q.Enqueue(new pair(i, j));
}
}
}
// Intialise step variable as zero
int step = 0;
// BFS approach
while (q.Count != 0)
{
int qsize = q.Count;
// Visit all indices
// with 1 and fill
// its neighbouring
// cells by 1
while (qsize-- > 0)
{
pair p = q.Peek();
q.Dequeue();
int i = p.first;
int j = p.second;
// Convert the neighbour
// from '0' to '1'
if ((j > 0) &&
mtrx[i, j - 1] == 0)
{
mtrx[i, j - 1] = 1;
q.Enqueue(new pair(i, j - 1));
}
// Convert the neighbour
// from '0' to '1'
if ((i < n - 1) &&
mtrx[i + 1, j] == 0)
{
mtrx[i + 1, j] = 1;
q.Enqueue(new pair(i + 1, j));
}
// Convert the neighbour
// from '0' to '1'
if ((j < n - 1) &&
mtrx[i, j + 1] == 0)
{
mtrx[i, j + 1] = 1;
q.Enqueue(new pair(i, j + 1));
}
// Convert the neighbour
// from '0' to '1'
if ((i > 0) &&
mtrx[i - 1, j] == 0)
{
mtrx[i - 1, j] = 1;
q.Enqueue(new pair(i - 1, j));
}
// Convert the neighbour
// from '0' to '1'
if ((i > 0) && (j > 0) &&
mtrx[i - 1, j - 1] == 0)
{
mtrx[i - 1, j - 1] = 1;
q.Enqueue(new pair(i - 1, j - 1));
}
// Convert the neighbour
// from '0' to '1'
if ((i > 0) && (j < (n - 1)) &&
mtrx[i - 1, j + 1] == 0)
{
mtrx[i - 1, j + 1] = 1;
q.Enqueue(new pair(i - 1, j + 1));
}
// Convert the neighbour
// from '0' to '1'
if ((i < (n - 1)) && (j < (n - 1)) &&
mtrx[i + 1, j + 1] == 0)
{
mtrx[i + 1, j + 1] = 1;
q.Enqueue(new pair(i + 1, j + 1));
}
// Convert the neighbour
// from '0' to '1'
if ((i < (n - 1)) && (j > 0) &&
mtrx[i + 1, j - 1] == 0)
{
mtrx[i + 1, j - 1] = 1;
q.Enqueue(new pair(i + 1, j - 1));
}
}
// Count steps
step++;
}
return step - 1;
}
// Driver code
public static void Main(String[] args)
{
// Dimension of matrix NXN
int n = 5;
int [,]mtrx = {{0, 0, 0, 0, 0},
{1, 0, 0, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 1, 0, 0},
{0, 0, 0, 1, 0}};
// Print number of steps
Console.Write(numberOfSteps(n, mtrx));
}
}
// This code is contributed by Rajput-Ji 输出：
3
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