给定一个r * c阶的二进制网格和一个初始位置。任务是找到从源到网格末端(第一行、最后一行、第一列或最后一列)的最小距离。只有当grid[i][j] = 0并且只允许向左、向右、向上和向下移动时,才能对单元格 grid[i][j] 进行移动。如果不存在有效路径,则打印-1 。
例子:
Input: i = 1, j = 1, grid[][] = { {1, 0, 1}, {0, 0, 0}, {1, 1, 1}}
Output: 1
Input: i = 0, j = 0, grid[][] = { {0, 1}, {1, 1}}
Output: 0
方法:
- 如果源已经是第一行/最后一行/列,则打印0 。
- 开始使用 BFS 从 source 开始遍历网格:
- 在队列中插入单元格位置。
- 从队列中弹出元素并将其标记为已访问。
- 对于与弹出的相邻的每个有效移动,将单元格位置插入队列。
- 在每次移动时,更新单元格与初始位置的最小距离。
- BFS 完成后,求第一行、最后一行、第一列和最后一列从源到每个单元格的最小距离。
- 最后打印其中的最小值。
下面是上述方法的实现:
CPP
// C++ implementation of the approach
#include
using namespace std;
#define row 5
#define col 5
// Global variables for grid, minDistance and visited array
int minDistance[row + 1][col + 1], visited[row + 1][col + 1];
// Queue for BFS
queue > que;
// Function to find whether the move is valid or not
bool isValid(int grid[][col], int i, int j)
{
if (i < 0 || j < 0
|| j >= col || i >= row
|| grid[i][j] || visited[i][j])
return false;
return true;
}
// Function to return the minimum distance
// from source to the end of the grid
int findMinPathminDistance(int grid[][col],
int sourceRow, int sourceCol)
{
// If source is one of the destinations
if (sourceCol == 0 || sourceCol == col - 1
|| sourceRow == 0 || sourceRow == row - 1)
return 0;
// Set minimum value
int minFromSource = row * col;
// Precalculate minDistance of each grid with R * C
for (int i = 0; i < row; i++)
for (int j = 0; j < col; j++)
minDistance[i][j] = row * col;
// Insert source position in queue
que.push(make_pair(sourceRow, sourceCol));
// Update minimum distance to visit source
minDistance[sourceRow][sourceCol] = 0;
// Set source to visited
visited[sourceRow][sourceCol] = 1;
// BFS approach for calculating the minDistance
// of each cell from source
while (!que.empty()) {
// Iterate over all four cells adjacent
// to current cell
pair cell = que.front();
// Initialize position of current cell
int cellRow = cell.first;
int cellCol = cell.second;
// Cell below the current cell
if (isValid(grid, cellRow + 1, cellCol)) {
// Push new cell to the queue
que.push(make_pair(cellRow + 1, cellCol));
// Update one of its neightbor's distance
minDistance[cellRow + 1][cellCol]
= min(minDistance[cellRow + 1][cellCol],
minDistance[cellRow][cellCol] + 1);
visited[cellRow + 1][cellCol] = 1;
}
// Above the current cell
if (isValid(grid, cellRow - 1, cellCol)) {
que.push(make_pair(cellRow - 1, cellCol));
minDistance[cellRow - 1][cellCol]
= min(minDistance[cellRow - 1][cellCol],
minDistance[cellRow][cellCol] + 1);
visited[cellRow - 1][cellCol] = 1;
}
// Right cell
if (isValid(grid, cellRow, cellCol + 1)) {
que.push(make_pair(cellRow, cellCol + 1));
minDistance[cellRow][cellCol + 1]
= min(minDistance[cellRow][cellCol + 1],
minDistance[cellRow][cellCol] + 1);
visited[cellRow][cellCol + 1] = 1;
}
// Left cell
if (isValid(grid, cellRow, cellCol - 1)) {
que.push(make_pair(cellRow, cellCol - 1));
minDistance[cellRow][cellCol - 1]
= min(minDistance[cellRow][cellCol - 1],
minDistance[cellRow][cellCol] + 1);
visited[cellRow][cellCol - 1] = 1;
}
// Pop the visited cell
que.pop();
}
int i;
// Minimum distance in the first row
for (i = 0; i < col; i++)
minFromSource = min(minFromSource, minDistance[0][i]);
// Minimum distance in the last row
for (i = 0; i < col; i++)
minFromSource = min(minFromSource, minDistance[row - 1][i]);
// Minimum distance in the first column
for (i = 0; i < row; i++)
minFromSource = min(minFromSource, minDistance[i][0]);
// Minimum distance in the last column
for (i = 0; i < row; i++)
minFromSource = min(minFromSource, minDistance[i][col - 1]);
// If no path exists
if (minFromSource == row * col)
return -1;
// Return the minimum distance
return minFromSource;
}
// Driver code
int main()
{
int sourceRow = 3, sourceCol = 3;
int grid[row][col] = { 1, 1, 1, 1, 0,
0, 0, 1, 0, 1,
0, 0, 1, 0, 1,
1, 0, 0, 0, 1,
1, 1, 0, 1, 0 };
cout << findMinPathminDistance(grid, sourceRow, sourceCol);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Pair class
static class Pair
{
int first,second;
Pair(int a, int b)
{
first = a;
second = b;
}
}
static int row = 5;
static int col = 5;
// Global variables for grid, minDistance and visited array
static int minDistance[][] = new int[row + 1][col + 1],
visited[][] = new int[row + 1][col + 1];
// Queue for BFS
static Queue que=new LinkedList<>();
// Function to find whether the move is valid or not
static boolean isValid(int grid[][], int i, int j)
{
if (i < 0 || j < 0
|| j >= col || i >= row
|| grid[i][j] != 0 || visited[i][j] != 0)
return false;
return true;
}
// Function to return the minimum distance
// from source to the end of the grid
static int findMinPathminDistance(int grid[][],
int sourceRow, int sourceCol)
{
// If source is one of the destinations
if (sourceCol == 0 || sourceCol == col - 1
|| sourceRow == 0 || sourceRow == row - 1)
return 0;
// Set minimum value
int minFromSource = row * col;
// Precalculate minDistance of each grid with R * C
for (int i = 0; i < row; i++)
for (int j = 0; j < col; j++)
minDistance[i][j] = row * col;
// Insert source position in queue
que.add(new Pair(sourceRow, sourceCol));
// Update minimum distance to visit source
minDistance[sourceRow][sourceCol] = 0;
// Set source to visited
visited[sourceRow][sourceCol] = 1;
// BFS approach for calculating the minDistance
// of each cell from source
while (que.size() > 0)
{
// Iterate over all four cells adjacent
// to current cell
Pair cell = que.peek();
// Initialize position of current cell
int cellRow = cell.first;
int cellCol = cell.second;
// Cell below the current cell
if (isValid(grid, cellRow + 1, cellCol))
{
// add new cell to the queue
que.add(new Pair(cellRow + 1, cellCol));
// Update one of its neightbor's distance
minDistance[cellRow + 1][cellCol]
= Math.min(minDistance[cellRow + 1][cellCol],
minDistance[cellRow][cellCol] + 1);
visited[cellRow + 1][cellCol] = 1;
}
// Above the current cell
if (isValid(grid, cellRow - 1, cellCol))
{
que.add(new Pair(cellRow - 1, cellCol));
minDistance[cellRow - 1][cellCol]
= Math.min(minDistance[cellRow - 1][cellCol],
minDistance[cellRow][cellCol] + 1);
visited[cellRow - 1][cellCol] = 1;
}
// Right cell
if (isValid(grid, cellRow, cellCol + 1))
{
que.add(new Pair(cellRow, cellCol + 1));
minDistance[cellRow][cellCol + 1]
= Math.min(minDistance[cellRow][cellCol + 1],
minDistance[cellRow][cellCol] + 1);
visited[cellRow][cellCol + 1] = 1;
}
// Left cell
if (isValid(grid, cellRow, cellCol - 1))
{
que.add(new Pair(cellRow, cellCol - 1));
minDistance[cellRow][cellCol - 1]
= Math.min(minDistance[cellRow][cellCol - 1],
minDistance[cellRow][cellCol] + 1);
visited[cellRow][cellCol - 1] = 1;
}
// Pop the visited cell
que.remove();
}
int i;
// Minimum distance in the first row
for (i = 0; i < col; i++)
minFromSource = Math.min(minFromSource,
minDistance[0][i]);
// Minimum distance in the last row
for (i = 0; i < col; i++)
minFromSource = Math.min(minFromSource,
minDistance[row - 1][i]);
// Minimum distance in the first column
for (i = 0; i < row; i++)
minFromSource = Math.min(minFromSource,
minDistance[i][0]);
// Minimum distance in the last column
for (i = 0; i < row; i++)
minFromSource = Math.min(minFromSource,
minDistance[i][col - 1]);
// If no path exists
if (minFromSource == row * col)
return -1;
// Return the minimum distance
return minFromSource;
}
// Driver code
public static void main(String args[])
{
int sourceRow = 3, sourceCol = 3;
int grid[][] = { {1, 1, 1, 1, 0},
{0, 0, 1, 0, 1},
{0, 0, 1, 0, 1},
{1, 0, 0, 0, 1},
{1, 1, 0, 1, 0 }};
System.out.println(findMinPathminDistance(grid,
sourceRow, sourceCol));
}
}
// This code is contributed by Arnab Kundu
Python
# Python3 implementation of the approach
from collections import deque as queue
row = 5
col = 5
# Global variables for grid, minDistance and visited array
minDistance = [[0 for i in range(col + 1)] for i in range(row + 1)]
visited = [[0 for i in range(col + 1)] for i in range(row + 1)]
# Queue for BFS
que = queue()
# Function to find whether the move is valid or not
def isValid(grid, i, j):
if (i < 0 or j < 0
or j >= col or i >= row
or grid[i][j] or visited[i][j]):
return False
return True
# Function to return the minimum distance
# from source to the end of the grid
def findMinPathminDistance(grid,sourceRow, sourceCol):
# If source is one of the destinations
if (sourceCol == 0 or sourceCol == col - 1
or sourceRow == 0 or sourceRow == row - 1):
return 0
# Set minimum value
minFromSource = row * col
# Precalculate minDistance of each grid with R * C
for i in range(row):
for j in range(col):
minDistance[i][j] = row * col
# Insert source position in queue
que.appendleft([sourceRow, sourceCol])
# Update minimum distance to visit source
minDistance[sourceRow][sourceCol] = 0;
# Set source to visited
visited[sourceRow][sourceCol] = 1;
# BFS approach for calculating the minDistance
# of each cell from source
while (len(que) > 0):
# Iterate over all four cells adjacent
# to current cell
cell = que.pop()
# Initialize position of current cell
cellRow = cell[0]
cellCol = cell[1]
# Cell below the current cell
if (isValid(grid, cellRow + 1, cellCol)):
# Push new cell to the queue
que.appendleft([cellRow + 1, cellCol])
# Update one of its neightbor's distance
minDistance[cellRow + 1][cellCol] = min(minDistance[cellRow + 1][cellCol],
minDistance[cellRow][cellCol] + 1)
visited[cellRow + 1][cellCol] = 1
# Above the current cell
if (isValid(grid, cellRow - 1, cellCol)):
que.appendleft([cellRow - 1, cellCol])
minDistance[cellRow - 1][cellCol] = min(minDistance[cellRow - 1][cellCol],
minDistance[cellRow][cellCol] + 1)
visited[cellRow - 1][cellCol] = 1
# Right cell
if (isValid(grid, cellRow, cellCol + 1)):
que.appendleft([cellRow, cellCol + 1])
minDistance[cellRow][cellCol + 1] = min(minDistance[cellRow][cellCol + 1],
minDistance[cellRow][cellCol] + 1)
visited[cellRow][cellCol + 1] = 1;
# Left cell
if (isValid(grid, cellRow, cellCol - 1)):
que.appendleft([cellRow, cellCol - 1])
minDistance[cellRow][cellCol - 1] = min(minDistance[cellRow][cellCol - 1],
minDistance[cellRow][cellCol] + 1)
visited[cellRow][cellCol - 1] = 1
# Pop the visited cell
# Minimum distance in the first row
for i in range(col):
minFromSource = min(minFromSource, minDistance[0][i]);
# Minimum distance in the last row
for i in range(col):
minFromSource = min(minFromSource, minDistance[row - 1][i]);
# Minimum distance in the first column
for i in range(row):
minFromSource = min(minFromSource, minDistance[i][0]);
# Minimum distance in the last column
for i in range(row):
minFromSource = min(minFromSource, minDistance[i][col - 1]);
# If no path exists
if (minFromSource == row * col):
return -1
# Return the minimum distance
return minFromSource
# Driver code
sourceRow = 3
sourceCol = 3
grid= [[1, 1, 1, 1, 0],
[0, 0, 1, 0, 1],
[0, 0, 1, 0, 1],
[1, 0, 0, 0, 1],
[1, 1, 0, 1, 0]]
print(findMinPathminDistance(grid, sourceRow, sourceCol))
# This code is contributed by mohit kumar 29
Javascript
输出:
2
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