二进制矩阵中距任何0个像元最接近的1个像元的所有距离的最大值

给定大小为N * N的矩阵，该矩阵填充有1和0 ，任务是查找从0单元格到最接近的1单元格的最大距离。如果矩阵仅填充0或1，则返回-1。

注意：矩阵中仅允许水平和垂直移动。

例子：

Input: 
mat[][] = {{0, 1, 0},
           {0, 0, 1},
           {0, 0, 0}}
Output: 3
Explanation: 
Cell number (2, 0) is at the farthest
distance of 3 cells from both the
1-cells (0, 1) and (1, 2).

Input: 
mat[][] = {{1, 0, 0},
           {0, 0, 0},
           {0, 0, 0}}
Output: 4
Explanation: 
Cell number (2, 2) is at the farthest 
distance of 4 cells from the only 
1-cell (1, 1).

方法1：天真的方法
对于每个0单元，计算其与每个1单元的距离并存储最小值。所有这些最小距离中的最大值就是答案。

下面是上述方法的实现：

C++

// C++ Program to find the maximum
// distance from a 0-cell to a 1-cell
 
#include 
using namespace std;
 
int maxDistance(vector >& grid)
{
    vector > one;
 
    int M = grid.size();
    int N = grid[0].size();
    int ans = -1;
 
    for (int i = 0; i < M; ++i) {
        for (int j = 0; j < N; ++j) {
            if (grid[i][j] == 1)
                one.emplace_back(i, j);
        }
    }
 
    // If the matrix consists of only 0's
    // or only 1's
    if (one.empty() || M * N == one.size())
        return -1;
 
    for (int i = 0; i < M; ++i) {
        for (int j = 0; j < N; ++j) {
 
            if (grid[i][j] == 1)
                continue;
 
            // If it's a 0-cell
            int dist = INT_MAX;
            for (auto& p : one) {
 
                // calculate its distance
                // with every 1-cell
                int d = abs(p.first - i)
                        + abs(p.second - j);
 
                // Compare and store the minimum
                dist = min(dist, d);
 
                if (dist <= ans)
                    break;
            }
 
            // Compare ans store the maximum
            ans = max(ans, dist);
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    vector > arr
        = { { 0, 0, 1 },
            { 0, 0, 0 },
            { 0, 0, 0 } };
 
    cout << maxDistance(arr) << endl;
    return 0;
}

Java

// Java Program to find the maximum
// distance from a 0-cell to a 1-cell
  
 
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 maxDistance(int [][]grid)
{
    Vector one = new Vector();
  
    int M = grid.length;
    int N = grid[0].length;
    int ans = -1;
  
    for (int i = 0; i < M; ++i) {
        for (int j = 0; j < N; ++j) {
            if (grid[i][j] == 1)
                one.add(new pair(i, j));
        }
    }
  
    // If the matrix consists of only 0's
    // or only 1's
    if (one.isEmpty() || M * N == one.size())
        return -1;
  
    for (int i = 0; i < M; ++i) {
        for (int j = 0; j < N; ++j) {
  
            if (grid[i][j] == 1)
                continue;
  
            // If it's a 0-cell
            int dist = Integer.MAX_VALUE;
            for (pair p : one) {
  
                // calculate its distance
                // with every 1-cell
                int d = Math.abs(p.first - i)
                        + Math.abs(p.second - j);
  
                // Compare and store the minimum
                dist = Math.min(dist, d);
  
                if (dist <= ans)
                    break;
            }
  
            // Compare ans store the maximum
            ans = Math.max(ans, dist);
        }
    }
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int [][]arr
        = { { 0, 0, 1 },
            { 0, 0, 0 },
            { 0, 0, 0 } };
  
    System.out.print(maxDistance(arr) +"\n");
}
}
 
// This code contributed by Princi Singh

Python3

# Python3 program to find the maximum
# distance from a 0-cell to a 1-cell
def maxDistance(grid):
   
  one = []
  M = len(grid)
  N = len(grid[0])
  ans = -1
   
  for i in range(M):
    for j in range(N):
      if (grid[i][j] == 1):
        one.append([i, j])
         
  # If the matrix consists of only 0's
  # or only 1's
  if (one == [] or M * N == len(one)):
    return -1
 
  for i in range(M):
    for j in range(N):
      if (grid[i][j] == 1):
        continue
 
      # If it's a 0-cell
      dist = float('inf')
       
      for p in one:
         
        # Calculate its distance
        # with every 1-cell
        d = abs(p[0] - i) + abs(p[1] - j)
         
        # Compare and store the minimum
        dist = min(dist, d)
         
        if (dist <= ans):
          break
 
        # Compare ans store the maximum
        ans = max(ans, dist)
         
  return ans
 
# Driver code
arr = [ [ 0, 0, 1 ],
        [ 0, 0, 0 ],
        [ 0, 0, 0 ] ]
 
print(maxDistance(arr))
 
# This code is contributed by rohitsingh07052

C#

// C# program to find the maximum
// distance from a 0-cell to a 1-cell
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 maxDistance(int [,]grid)
{
    List one = new List();
 
    int M = grid.GetLength(0);
    int N = grid.GetLength(1);
    int ans = -1;
 
    for(int i = 0; i < M; ++i)
    {
       for(int j = 0; j < N; ++j)
       {
          if (grid[i, j] == 1)
              one.Add(new pair(i, j));
       }
    }
 
    // If the matrix consists of only 0's
    // or only 1's
    if (one.Count == 0 || M * N == one.Count)
        return -1;
 
    for(int i = 0; i < M; ++i)
    {
       for(int j = 0; j < N; ++j)
       {
          if (grid[i, j] == 1)
              continue;
           
          // If it's a 0-cell
          int dist = int.MaxValue;
          foreach (pair p in one)
          {
               
              // Calculate its distance
              // with every 1-cell
              int d = Math.Abs(p.first - i) +
                      Math.Abs(p.second - j);
               
              // Compare and store the minimum
              dist = Math.Min(dist, d);
               
              if (dist <= ans)
                  break;
          }
           
          // Compare ans store the maximum
          ans = Math.Max(ans, dist);
       }
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int [,]arr = { { 0, 0, 1 },
                   { 0, 0, 0 },
                   { 0, 0, 0 } };
 
    Console.Write(maxDistance(arr) + "\n");
}
}
 
// This code is contributed by Amit Katiyar

输出：

4

时间复杂度： O(M * N * P) ，其中网格的大小为M * N ， P为1个像元的计数。
辅助空间： O(P)

方法2：使用BFS
从1单元格开始，并逐层执行广度优先搜索遍历。我们可以找到的最大层数是我们的答案。

下面是上述方法的实现：

C++

// C++ Program to find the maximum
// distance from a 0-cell to a 1-cell
 
#include 
using namespace std;
 
// Function to find the maximum distance
int maxDistance(vector >& grid)
{
    // Queue to store all 1-cells
    queue > q;
 
    // Grid dimensions
    int M = grid.size();
    int N = grid[0].size();
    int ans = -1;
 
    // Directions traversable from
    // a given a particular cell
    int dirs[4][2] = { { 0, 1 },
                       { 1, 0 },
                       { 0, -1 },
                       { -1, 0 } };
 
    for (int i = 0; i < M; ++i) {
        for (int j = 0; j < N; ++j) {
            if (grid[i][j] == 1)
                q.emplace(i, j);
        }
    }
 
    // If the grid contains
    // only 0s or only 1s
    if (q.empty() || M * N == q.size())
        return -1;
 
    while (q.size()) {
 
        int cnt = q.size();
 
        while (cnt--) {
 
            // Access every 1-cell
            auto p = q.front();
            q.pop();
 
            // Traverse all possible
            // directions from the cells
            for (auto& dir : dirs) {
 
                int x = p.first + dir[0];
                int y = p.second + dir[1];
 
                // Check if the cell is
                // within the boundaries
                // or contains a 1
                if (x < 0 || x >= M
                    || y < 0 || y >= N
                    || grid[x][y])
                    continue;
 
                q.emplace(x, y);
                grid[x][y] = 1;
            }
        }
        ++ans;
    }
    return ans;
}
 
// Driver code
int main()
{
    vector > arr = { { 0, 0, 1 },
                                 { 0, 0, 0 },
                                 { 0, 0, 1 } };
 
    cout << maxDistance(arr) << endl;
    return 0;
}

Java

// Java program to find the maximum
// distance from a 0-cell to a 1-cell
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 find the maximum distance
static int maxDistance(int [][]grid)
{
     
    // Queue to store all 1-cells
    Queue q = new LinkedList();
 
    // Grid dimensions
    int M = grid.length;
    int N = grid[0].length;
    int ans = -1;
 
    // Directions traversable from
    // a given a particular cell
    int dirs[][] = { { 0, 1 },
                     { 1, 0 },
                     { 0, -1 },
                     { -1, 0 } };
 
    for(int i = 0; i < M; ++i)
    {
        for(int j = 0; j < N; ++j)
        {
            if (grid[i][j] == 1)
                q.add(new pair(i, j));
        }
    }
 
    // If the grid contains
    // only 0s or only 1s
    if (q.isEmpty() || M * N == q.size())
        return -1;
 
    while (q.size() > 0)
    {
        int cnt = q.size();
        while (cnt-->0)
        {
 
            // Access every 1-cell
            pair p = q.peek();
            q.remove();
 
            // Traverse all possible
            // directions from the cells
            for(int []dir : dirs)
            {
                int x = p.first + dir[0];
                int y = p.second + dir[1];
 
                // Check if the cell is
                // within the boundaries
                // or contains a 1
                if (x < 0 || x >= M ||
                    y < 0 || y >= N ||
                    grid[x][y] > 0)
                    continue;
 
                q.add(new pair(x, y));
                grid[x][y] = 1;
            }
        }
        ++ans;
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int [][]arr = { { 0, 0, 1 },
                    { 0, 0, 0 },
                    { 0, 0, 1 } };
 
    System.out.print(maxDistance(arr) + "\n");
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to find the maximum
# distance from a 0-cell to a 1-cell
 
# Function to find the maximum distance
def maxDistance(grid):
 
    # Queue to store all 1-cells
    q = []
  
    # Grid dimensions
    M = len(grid)
    N = len(grid[0])
    ans = -1
  
    # Directions traversable from
    # a given a particular cell
    dirs = [ [ 0, 1 ], [ 1, 0 ],
             [ 0, -1 ], [ -1, 0 ] ]
  
    for i in range(M):
        for j in range(N):
            if (grid[i][j] == 1):
                q.append([i, j])
  
    # If the grid contains
    # only 0s or only 1s
    if (len(q) == 0 or M * N == len(q)):
        return -1
  
    while (len(q) > 0):
        cnt = len(q)
  
        while (cnt > 0):
  
            # Access every 1-cell
            p = q[0]
            q.pop()
  
            # Traverse all possible
            # directions from the cells
            for Dir in dirs:
                x = p[0] + Dir[0]
                y = p[1] + Dir[1]
  
                # Check if the cell is
                # within the boundaries
                # or contains a 1
                if (x < 0 or x >= M or
                    y < 0 or y >= N or
                    grid[x][y]):
                    continue
  
                q.append([x, y])
                grid[x][y] = 1
                 
            cnt -= 1
             
        ans += 2
     
    return ans
     
# Driver code   
arr = [ [ 0, 0, 1 ],
        [ 0, 0, 0 ],
        [ 0, 0, 1 ] ]
         
print(maxDistance(arr))
 
# This code is contributed by divyeshrabadiya07

C＃

// C# program to find
// the maximum distance
// from a 0-cell to a 1-cell
using System;
using System.Collections.Generic;
class GFG{
     
static int index = 0;
class pair
{
  public int first, second;
  public pair(int first, int second)
  {
    this.first = first;
    this.second = second;
  }
}
 
// Function to find the
// maximum distance
static int maxDistance(int [,]grid)
{
  // Queue to store all 1-cells
  Queue q = new Queue();
 
  // Grid dimensions
  int M = grid.GetLength(0);
  int N = grid.GetLength(1);
  int ans = -1;
 
  // Directions traversable from
  // a given a particular cell
  int [,]dirs = {{0, 1},
                 {1, 0},
                 {0, -1},
                 {-1, 0}};
 
  for(int i = 0; i < M; ++i)
  {
    for(int j = 0; j < N; ++j)
    {
      if (grid[i, j] == 1)
        q.Enqueue(new pair(i, j));
    }
  }
 
  // If the grid contains
  // only 0s or only 1s
  if (q.Count==0 || M * N == q.Count)
    return -1;
 
  while (q.Count > 0)
  {
    int cnt = q.Count;
    while (cnt-- > 0)
    {
      // Access every 1-cell
      pair p = q.Peek();
      q.Dequeue();
 
      // Traverse all possible
      // directions from the cells
 
      for(int i = 0; i < dirs.GetLength(0);)
      {
        int []dir = GetRow(dirs, i++);
        int x = p.first + dir[0];
        int y = p.second + dir[1];
 
        // Check if the cell is
        // within the boundaries
        // or contains a 1
        if (x < 0 || x >= M ||
            y < 0 || y >= N ||
            grid[x, y] > 0)
          continue;
 
        q.Enqueue(new pair(x, y));
        grid[x, y] = 1;
      }
    }
    ++ans;
  }
  return ans;
}
   
public static int[] GetRow(int[,] matrix,
                           int row)
{
  var rowLength = matrix.GetLength(1);
  var rowVector = new int[rowLength];
 
  for (var i = 0; i < rowLength; i++)
    rowVector[i] = matrix[row, i];
  return rowVector;
}
   
// Driver code
public static void Main(String[] args)
{
  int [,]arr = {{0, 0, 1},
                {0, 0, 0},
                {0, 0, 1}};
  Console.Write(maxDistance(arr) + "\n");
}
}
 
// This code is contributed by shikhasingrajput

输出：

3

时间复杂度： O(M * N)
辅助空间： O(M * N)

方法3：使用动态编程

• 不断更新已行驶的最大距离的矩阵。
• 从矩阵的左上角单元格(0，0)遍历到右下角。令grid [i] [j]代表距左侧或上方最近的1个像元(或者当然是其自身)的最大距离。
• 从右下到左上进行第二遍，更新网格数组，将单元格grid [i] [j]定义为grid [i] [j] ， grid [i + 1] [j]和grid的最小值[i] [j + 1] 。
• 在从右下到左上的遍历过程中跟踪最大值，并在末尾返回该值。如果值为0，即网格仅填充0或仅填充1，则返回-1。

下面是上述方法的实现：

C++

// C++ Program to find the maximum
// distance from a 0-cell to a 1-cell
 
#include 
using namespace std;
 
// Function to find the maximum distance
int maxDistance(vector >& grid)
{
    if (!grid.size())
        return -1;
    int N = grid.size();
 
    int INF = 1000000;
 
    // DP matrix
    vector >
    dp(N, vector(N, 0));
 
    grid[0][0] = grid[0][0] == 1
                     ? 0
                     : INF;
 
    // Set up top row and left column
    for (int i = 1; i < N; i++)
        grid[0][i] = grid[0][i] == 1
                         ? 0
                         : grid[0][i - 1] + 1;
    for (int i = 1; i < N; i++)
        grid[i][0] = grid[i][0] == 1
                         ? 0
                         : grid[i - 1][0] + 1;
 
    // Pass one: top left to bottom right
    for (int i = 1; i < N; i++) {
        for (int j = 1; j < N; j++) {
            grid[i][j] = grid[i][j] == 1
                             ? 0
                             : min(grid[i - 1][j],
                                   grid[i][j - 1])
                                   + 1;
        }
    }
 
    // Check if there was no "One" Cell
    if (grid[N - 1][N - 1] >= INF)
        return -1;
 
    // Set up top row and left column
    int maxi = grid[N - 1][N - 1];
    for (int i = N - 2; i >= 0; i--) {
        grid[N - 1][i]
            = min(grid[N - 1][i],
                  grid[N - 1][i + 1] + 1);
        maxi = max(grid[N - 1][i], maxi);
    }
 
    for (int i = N - 2; i >= 0; i--) {
        grid[i][N - 1]
            = min(grid[i][N - 1],
                  grid[i + 1][N - 1] + 1);
        maxi = max(grid[i][N - 1], maxi);
    }
 
    // Past two: bottom right to top left
    for (int i = N - 2; i >= 0; i--) {
        for (int j = N - 2; j >= 0; j--) {
            grid[i][j] = min(grid[i][j],
                             min(grid[i + 1][j] + 1,
                                 grid[i][j + 1] + 1));
            maxi = max(grid[i][j], maxi);
        }
    }
 
    return !maxi ? -1 : maxi;
}
 
// Driver code
int main()
{
    vector > arr = { { 0, 0, 1 },
                                 { 0, 0, 0 },
                                 { 0, 0, 0 } };
 
    cout << maxDistance(arr) << endl;
    return 0;
}

Java

// Java program to find the maximum
// distance from a 0-cell to a 1-cell
import java.util.*;
import java.lang.*;
 
class GFG{
     
// Function to find the maximum distance
static int maxDistance(int[][] grid)
{
    if (grid.length == 0)
        return -1;
         
    int N = grid.length;
    int INF = 1000000;
     
    grid[0][0] = grid[0][0] == 1 ? 0 : INF;
     
    // Set up top row and left column
    for(int i = 1; i < N; i++)
        grid[0][i] = grid[0][i] == 1 ? 0 :
                     grid[0][i - 1] + 1;
                      
    for(int i = 1; i < N; i++)
        grid[i][0] = grid[i][0] == 1 ? 0 :
                     grid[i - 1][0] + 1;
  
    // Pass one: top left to bottom right
    for(int i = 1; i < N; i++)
    {
        for(int j = 1; j < N; j++)
        {
            grid[i][j] = grid[i][j] == 1 ? 0 :
                Math.min(grid[i - 1][j],
                         grid[i][j - 1]) + 1;
        }
    }
  
    // Check if there was no "One" Cell
    if (grid[N - 1][N - 1] >= INF)
        return -1;
  
    // Set up top row and left column
    int maxi = grid[N - 1][N - 1];
    for(int i = N - 2; i >= 0; i--)
    {
        grid[N - 1][i] = Math.min(
            grid[N - 1][i],
            grid[N - 1][i + 1] + 1);
             
        maxi = Math.max(grid[N - 1][i], maxi);
    }
  
    for(int i = N - 2; i >= 0; i--)
    {
        grid[i][N - 1] = Math.min(
            grid[i][N - 1],
            grid[i + 1][N - 1] + 1);
             
        maxi = Math.max(grid[i][N - 1], maxi);
    }
  
    // Past two: bottom right to top left
    for(int i = N - 2; i >= 0; i--)
    {
        for(int j = N - 2; j >= 0; j--)
        {
            grid[i][j] = Math.min(
                grid[i][j],
                Math.min(grid[i + 1][j] + 1,
                         grid[i][j + 1] + 1));
                          
            maxi = Math.max(grid[i][j], maxi);
        }
    }
  
    return maxi == 0 ? -1 : maxi;
}
 
// Driver code
public static void main(String[] args)
{
    int[][] arr = { { 0, 0, 1 },
                    { 0, 0, 0 },
                    { 0, 0, 0 } };
     
    System.out.println(maxDistance(arr));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to find the maximum
# distance from a 0-cell to a 1-cell
 
# Function to find the maximum distance
def maxDistance(grid):
     
    if (len(grid) == 0):
        return -1
          
    N = len(grid)
    INF = 1000000
     
    if grid[0][0] == 1:
        grid[0][0] = 0
    else:
        grid[0][0] = INF
      
    # Set up top row and left column
    for i in range(1, N):
        if grid[0][i] == 1:
            grid[0][i] = 0
        else:
            grid[0][i] = grid[0][i - 1] + 1
                       
    for i in range(1, N):
        if grid[i][0] == 1:
            grid[i][0] = 0
        else:
            grid[i][0] = grid[i - 1][0] + 1
   
    # Pass one: top left to bottom right
    for i in range(1, N):
        for j in range(1, N):
            if grid[i][j] == 1:
                grid[i][j] = 0
            else:
                grid[i][j] = min(grid[i - 1][j],
                                 grid[i][j - 1] + 1)
   
    # Check if there was no "One" Cell
    if (grid[N - 1][N - 1] >= INF):
        return -1
   
    # Set up top row and left column
    maxi = grid[N - 1][N - 1]
     
    for i in range(N - 2, -1, -1):
        grid[N - 1][i] = min(grid[N - 1][i],
                             grid[N - 1][i + 1] + 1)
              
        maxi = max(grid[N - 1][i], maxi)
   
    for i in range(N - 2, -1, -1):
        grid[i][N - 1] = min(grid[i][N - 1],
                             grid[i + 1][N - 1] + 1)
              
        maxi = max(grid[i][N - 1], maxi + 1)
   
    # Past two: bottom right to top left
    for i in range(N - 2, -1, -1):
        for j in range(N - 2, -1, -1):
            grid[i][j] = min(grid[i][j],
                             min(grid[i + 1][j] + 1,
                                 grid[i][j + 1] + 1))
                           
            maxi = max(grid[i][j], maxi)
             
    if maxi == 0:
        return -1
    else:
        return maxi
 
# Driver code       
arr = [ [ 0, 0, 1 ], [ 0, 0, 0 ], [ 0, 0, 0 ] ]
      
print(maxDistance(arr))
 
# This code is contributed by divyesh072019

C＃

// C# program to find the maximum
// distance from a 0-cell to a 1-cell
using System;
class GFG {
 
    // Function to find the maximum distance
    static int maxDistance(int[, ] grid)
    {
        if (grid.GetLength(0) == 0)
            return -1;
 
        int N = grid.GetLength(0);
        int INF = 1000000;
 
        grid[0, 0] = grid[0, 0] == 1 ? 0 : INF;
 
        // Set up top row and left column
        for (int i = 1; i < N; i++)
            grid[0, i]
                = grid[0, i] == 1 ? 0 : grid[0, i - 1] + 1;
 
        for (int i = 1; i < N; i++)
            grid[i, 0]
                = grid[i, 0] == 1 ? 0 : grid[i - 1, 0] + 1;
 
        // Pass one: top left to bottom right
        for (int i = 1; i < N; i++) {
            for (int j = 1; j < N; j++) {
                grid[i, j] = grid[i, j] == 1
                                 ? 0
                                 : Math.Min(grid[i - 1, j],
                                            grid[i, j - 1])
                                       + 1;
            }
        }
 
        // Check if there was no "One" Cell
        if (grid[N - 1, N - 1] >= INF)
            return -1;
 
        // Set up top row and left column
        int maxi = grid[N - 1, N - 1];
        for (int i = N - 2; i >= 0; i--) {
            grid[N - 1, i] = Math.Min(
                grid[N - 1, i], grid[N - 1, i + 1] + 1);
 
            maxi = Math.Max(grid[N - 1, i], maxi);
        }
 
        for (int i = N - 2; i >= 0; i--) {
            grid[i, N - 1] = Math.Min(
                grid[i, N - 1], grid[i + 1, N - 1] + 1);
 
            maxi = Math.Max(grid[i, N - 1], maxi);
        }
 
        // Past two: bottom right to top left
        for (int i = N - 2; i >= 0; i--) {
            for (int j = N - 2; j >= 0; j--) {
                grid[i, j] = Math.Min(
                    grid[i, j],
                    Math.Min(grid[i + 1, j] + 1,
                             grid[i, j + 1] + 1));
 
                maxi = Math.Max(grid[i, j], maxi);
            }
        }
 
        return maxi == 0 ? -1 : maxi;
    }
 
    // Driver code
    public static void Main()
    {
        int[, ] arr
            = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 0 } };
 
        Console.WriteLine(maxDistance(arr));
    }
}
 
// This code is contributed by subhammahato348

输出

4

时间复杂度： O(M * N)
辅助空间： O(1)

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