问题描述
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“ s”是起点。并且有多个目的地点“ x”。我只需要找到一个目的地。如果目的地接近起点,则BFS算法可以很快找到解决方案。如果它们像上面的示例中那样离得很远,则将花费无尽的时间。 所以我的问题是: a)该算法对这种特定类型的迷宫不利吗?我应该使用A *还是类似的东西? b)我的实现不好吗?
实施:
public class BFS {
public static String getPath(String[][] map) {
String[] ways = { "L","R","U","D" }; // directions to go
Queue<String> q = new LinkedList<>();
q.offer("");
String path = "";
while (!foundBait(map,path)) {
path = q.poll();
for (String s : ways) {
String newPath = path + s;
if (valid(map,newPath))
q.offer(newPath);
}
}
return path;
}
private static boolean foundBait(String[][] map,String moves) {
int xStart = 0;
int yStart = 0;
for (int y = 0; y < map.length; y++)
for (int x = 0; x < map[0].length; x++)
if (map[y][x].equals("s")) {
xStart = x;
yStart = y;
}
int i = xStart;
int j = yStart;
for (int s = 0; s < moves.length(); s++) {
if (moves.charat(s) == "L".charat(0))
i--;
else if (moves.charat(s) == "R".charat(0))
i++;
else if (moves.charat(s) == "U".charat(0))
j--;
else if (moves.charat(s) == "D".charat(0))
j++;
}
if (map[j][i].equals("x"))
return true;
return false;
}
private static boolean valid(String[][] map,String moves) {
int xStart = 0;
int yStart = 0;
for (int y = 0; y < map.length; y++)
for (int x = 0; x < map[0].length; x++)
if (map[y][x].equals("s")) {
xStart = x;
yStart = y;
}
int i = xStart;
int j = yStart;
for (int s = 0; s < moves.length(); s++) {
if (moves.charat(s) == "L".charat(0))
i--;
else if (moves.charat(s) == "R".charat(0))
i++;
else if (moves.charat(s) == "U".charat(0))
j--;
else if (moves.charat(s) == "D".charat(0))
j++;
if (!(0 <= i && i < map[0].length && 0 <= j && j < map.length))
return false;
else if (map[j][i].equals("#") || map[j][i].equals("-"))
return false;
}
return true;
}
}
解决方法
如评论中所述,问题不是标记添加到路径的节点,解决方案是使用第二个矩阵进行标记。