二进制矩阵中从任何 0 个单元到最近的 1 个单元的所有距离的最大值
原文:https://www . geeksforgeeks . org/find-从二进制矩阵中的任意 0 个单元到最近的 1 个单元的最大距离/
给定一个大小为 N*N 的矩阵，其中填充了 1 和 0 ，任务是找到从 0 像元到最近的 1 像元的最大距离。如果矩阵只有 0 或 1，返回-1。
注意:矩阵中只允许水平和垂直移动。
示例:
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，计算它与每个 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 <bits/stdc++.h>
using namespace std;
int maxDistance(vector<vector<int> >& grid)
{
vector<pair<int, int> > 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 and store the maximum
ans = max(ans, dist);
}
}
return ans;
}
// Driver code
int main()
{
vector<vector<int> > arr
= { { 0, 0, 1 },
{ 0, 0, 0 },
{ 0, 0, 0 } };
cout << maxDistance(arr) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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<pair> one = new Vector<pair>();
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 and 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
Python 3
# 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 and 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<pair> one = new List<pair>();
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 and 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
Output:
4
时间复杂度: O(MNP) 其中网格大小为M * NP为 1-细胞计数。 辅助空间: O(P)
方法 2:使用【BFS】 从一个 1 单元开始，执行广度优先搜索遍历，逐层进行。我们能检索到的最大层数就是我们的答案。
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 <bits/stdc++.h>
using namespace std;
// Function to find the maximum distance
int maxDistance(vector<vector<int> >& grid)
{
// Queue to store all 1-cells
queue<pair<int, int> > 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<vector<int> > arr = { { 0, 0, 1 },
{ 0, 0, 0 },
{ 0, 0, 1 } };
cout << maxDistance(arr) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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<pair> q = new LinkedList<pair>();
// 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
Python 3
# 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<pair> q = new Queue<pair>();
// 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
Output:
3
时间复杂度:O(M * N) T5】辅助空间: O(MN)*
方法 3:使用 动态编程
- 不断更新已行驶的最大距离矩阵。
- 从矩阵左上角单元格 (0，0) 向右下角遍历。让网格[i][j]代表离左边或上面最近的 1-cell 的最大距离(当然也包括它本身)。
- 从右下到左上进行第二遍，更新网格数组，将单元格网格【I】【j】定义为网格【I】【j】网格【I+1】【j】和网格【I】【j+1】的最小值。
- 在从右下到左上的遍历过程中跟踪最大值，并在最后返回该值。如果该值为 0，即网格仅填充 0 或 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 <bits/stdc++.h>
using namespace std;
// Function to find the maximum distance
int maxDistance(vector<vector<int> >& grid)
{
if (!grid.size())
return -1;
int N = grid.size();
int INF = 1000000;
// DP matrix
vector<vector<int> >
dp(N, vector<int>(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<vector<int> > arr = { { 0, 0, 1 },
{ 0, 0, 0 },
{ 0, 0, 0 } };
cout << maxDistance(arr) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
Output
4
时间复杂度:O(M * N) T5】辅助空间: O(1)
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