我实现了一条计算最短路线的代码,但我想添加一个条件。如果我得到这样的矩阵:
1 0 0 0 0 0 0 0
1 0 0 1 1 1 1 0
0 0 0 0 0 0 0 0
0 0 1 1 1 0 1 0
A 0 0 0 0 0 0 0
0 0 1 1 0 B 0 1
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0
我如何添加一个条件,即从单元格A到单元格B只有在行距为1或一列为0的情况下才能移动,因此从A到B的最短路径(A在[4; 0]和B上在[5; 5]上)是:[4; 0],[5; 0],[6; 0],[7; 0],[7; 1],[7; 2],[7,3 ],[7,4],[7,5],[6,5],[5; 5]。
这是我目前所拥有的:
import java.util.LinkedList;
public class ShortestPathBetweenCellsBFS {
private static class Cell {
int x;
int y;
int dist; //distance
Cell prev; //parent cell in the path
Cell(int x, int y, int dist, Cell prev) {
this.x = x;
this.y = y;
this.dist = dist;
this.prev = prev;
}
@Override
public String toString(){
return "(" + x + "," + y + ")";
}
}
public static void shortestPath(int[][] matrix, int[] start, int[] end) {
int sx = start[0], sy = start[1];
int dx = end[0], dy = end[1];
int m = matrix.length;
int n = matrix[0].length;
Cell[][] cells = new Cell[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] != 1) {
cells[i][j] = new Cell(i, j, Integer.MAX_VALUE, null);
}
}
}
LinkedList<Cell> queue = new LinkedList<>();
Cell src = cells[sx][sy];
src.dist = 0;
queue.add(src);
System.out.println(src);
Cell dest = null;
Cell p;
while ((p = queue.poll()) != null) {
//find destination
if (p.x == dx && p.y == dy) {
dest = p;
break;
}
// moving up
visit(cells, queue, p.x - 1, p.y, p);
// moving down
visit(cells, queue, p.x + 1, p.y, p);
// moving left
visit(cells, queue, p.x, p.y - 1, p);
//moving right
visit(cells, queue, p.x, p.y + 1, p);
}
if (dest == null) {
return;
} else {
LinkedList<Cell> path = new LinkedList<>();
p = dest;
do {
path.addFirst(p);
} while ((p = p.prev) != null);
System.out.println(path);
}
}
//function to update cell visiting status
static void visit(Cell[][] cells, LinkedList<Cell> queue, int x, int y, Cell parent) {
if (x < 0 || x >= cells.length || y < 0 || y >= cells[0].length || cells[x][y] == null) {
return;
}
int dist = parent.dist + 1;
Cell p = cells[x][y];
if (dist < p.dist) {
p.dist = dist;
p.prev = parent;
queue.add(p);
}
}
public static void main(String[] args) {
int[][] matrix = {
{1, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 1, 1, 1, 1, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 1, 1, 1, 0, 1, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 1, 1, 0, 0, 0, 1},
{0, 0, 0, 0, 0, 0, 0, 1},
{0, 0, 0, 0, 0, 0, 0, 0}
};
int[] start = {4, 0};
int[] end = {5, 5};
shortestPath(matrix, start, end);
}
}
最佳答案
一种可能的解决方案的基本思想是,如果到任何墙壁的距离至少为一个,则仅将单元格添加到寻路算法的队列中。
因此,例如在您的visit
方法中,您还必须检查所有相邻的单元格,而不是仅检查cells[x][y]
。
一种方法是在发现障碍物的单元格上遍历相邻单元格并中止。
for (int x_test = Math.max(0, x - 1); x_test <= Math.min(cells.length - 1; x + 1); x_test++) {
for (int y_test = Math.max(0, y - 1); y_test <= Math.min(cells[0].length - 1; y + 1); y_test++) {
if (cells[x][y] == null) {
return;
}
}
}
请注意,我在循环中使用
Math.min
和Math.max
来防止超出范围的访问。