从源
到网格角的最小距离
原文： https://www.geeksforgeeks.org/minimum-distance-to-the-corner-of-a-grid-from-source/
给定二阶网格 r * c 和初始位置。 任务是找到从源到网格任何角落的最小距离。 仅当 grid [i] [j] = 0 并且仅剩下，时，才能移动到单元格 grid [i] [j] 允许向右，向上和向下移动。 如果没有有效的路径，则打印-1。
示例：
输入：i = 1，j = 1，grid [] [] = {{0，0，1}，{0，0，0}，{1，1，1}} 输出：2 （1、1）->（1、0）->（0、0）
输入：i = 0，j = 0，grid [] [] = {{0，1}，{1，1}} 输出：0 源已经是网格的一角。
方法：
-
如果来源已经在角落，则打印
0。 -
使用 BFS 从源开始遍历网格：
-
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。
-
对于与弹出的动作相邻的每个有效动作，将单元格位置插入队列。
-
每次移动时，更新单元格到初始位置的最小距离。
-
-
BFS 完成后，找到从源到每个角的最小距离。
-
最后打印其中的最小值。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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<pair<int, int> > 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 minDistance(int grid[][col],
int sourceRow, int sourceCol)
{
// If source is one of the destinations
if ((sourceCol == 0 && sourceRow == 0)
|| (sourceCol == col - 1 && sourceRow == 0)
|| (sourceCol == 0 && sourceRow == row - 1)
|| (sourceCol == col - 1 && 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<int, int> 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 to the corner
// of the first row, first column
minFromSource = min(minFromSource,
minDistance[0][0]);
// Minimum distance to the corner
// of the last row, first column
minFromSource = min(minFromSource,
minDistance[row - 1][0]);
// Minimum distance to the corner
// of the last row, last column
minFromSource = min(minFromSource,
minDistance[row - 1][col - 1]);
// Minimum distance to the corner
// of the first row, last column
minFromSource = min(minFromSource,
minDistance[0][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, 0, 0,
0, 0, 1, 0, 1,
0, 0, 1, 0, 1,
1, 0, 0, 0, 1,
1, 1, 0, 1, 0 };
cout << minDistance(grid, sourceRow, sourceCol);
return 0;
}
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