我有一个格子[40 x 15]，上面有2到16个单位，以及未知数量的障碍物 如何从我的单位位置找到所有单位的最短路径。
我有两个辅助方法，我们可以考虑为O（1）
我实施了A* search算法，可同时搜索所有路线。最后，它输出一个网格，其中每个单元格都有一个数字，表示距离我的位置的距离，以及网格上所有单元的集合。它用O（N）执行，其中N是单元格的数量 - 在我的情况下是600。
我使用AS3实现这一点，不幸的是，我的机器需要30-50毫秒来计算。
这是我的源代码。你能建议我一个更好的方法吗？
package com.gazman.strategy_of_battle_package.map
{
import flash.geom.Point;
/**
* Implementing a path finding algorithm(Similar to A* search only there is no known target) to calculate the shortest path to each cell on the map.
* Once calculation is complete the information will be available at cellsMap. Each cell is a number representing the
* number of steps required to get to that location. Enemies and Allies will be represented with negative distance. Also the enemy and Allys
* coordinations collections are provided. Blocked cells will have the value 0.<br><br>
* Worth case and best case efficiency is O(N) where N is the number of cells.
*/
public class MapFilter
{
private static const PULL:Vector.<MapFilter> = new Vector.<MapFilter>();
public var cellsMap:Vector.<Vector.<int>>;
public var allys:Vector.<Point>;
public var enemies:Vector.<Point>;
private var stack:Vector.<MapFilter>;
private var map:Map;
private var x:int;
private var y:int;
private var count:int;
private var commander:String;
private var hash:Object;
private var filtered:Boolean;
public function filter(map:Map, myLocation:Point, commander:String):void{
filtered = true;
this.commander = commander;
this.map = map;
this.x = myLocation.x;
this.y = myLocation.y;
init();
cellsMap[x][y] = 1;
excecute();
while(stack.length > 0){
var length:int = stack.length;
for(var i:int = 0; i < length; i++){
var mapFilter:MapFilter = stack.shift();
mapFilter.excecute();
PULL.push(mapFilter);
}
}
}
public function navigateTo(location:Point):Point{
if(!filtered){
throw new Error("Must filter before navigating");
}
var position:int = Math.abs(cellsMap[location.x][location.y]);
if(position == 0){
throw new Error("Target unreachable");
}
while(position > 2){
if(canNavigateTo(position, location.x + 1, location.y)){
location.x++;
}
else if(canNavigateTo(position, location.x - 1, location.y)){
location.x--;
}
else if(canNavigateTo(position, location.x, location.y + 1)){
location.y++;
}
else if(canNavigateTo(position, location.x, location.y - 1)){
location.y--;
}
position = cellsMap[location.x][location.y];
}
return location;
throw new Error("Unexpected filtering error");
}
private function canNavigateTo(position:int, targetX:int, targetY:int):Boolean
{
return isInMapRange(targetX, targetY) && cellsMap[targetX][targetY] < position && cellsMap[targetX][targetY] > 0;
}
private function excecute():void
{
papulate(x + 1, y);
papulate(x - 1, y);
papulate(x, y + 1);
papulate(x, y - 1);
}
private function isInMapRange(x:int, y:int):Boolean{
return x < cellsMap.length &&
x >= 0 &&
y < cellsMap[0].length &&
y >= 0;
}
private function papulate(x:int, y:int):void
{
if(!isInMapRange(x,y) ||
cellsMap[x][y] != 0 ||
hash[x + "," + y] != null ||
map.isBlocked(x,y)){
return;
}
// we already checked that is not block
// checking if there units
if(map.isEmpty(x,y)){
cellsMap[x][y] = count;
addTask(x,y);
}
else{
cellsMap[x][y] = -count;
if(map.isAlly(x,y, commander)){
hash[x + "," + y] = true;
allys.push(new Point(x,y));
}
else {
hash[x + "," + y] = true;
enemies.push(new Point(x,y));
}
}
}
private function addTask(x:int, y:int):void
{
var mapFilter:MapFilter = PULL.pop();
if(mapFilter == null){
mapFilter = new MapFilter();
}
mapFilter.commander = commander;
mapFilter.hash = hash;
mapFilter.map = map;
mapFilter.cellsMap = cellsMap;
mapFilter.allys = allys;
mapFilter..enemies = enemies;
mapFilter.stack = stack;
mapFilter.count = count + 1;
mapFilter.x = x;
mapFilter.y = y;
stack.push(mapFilter);
}
private function init():void
{
hash = new Object();
cellsMap = new Vector.<Vector.<int>>();
for(var i:int = 0; i < map.width;i++){
cellsMap.push(new Vector.<int>);
for(var j:int = 0; j < map.height;j++){
cellsMap[i].push(0);
}
}
allys = new Vector.<Point>();
enemies = new Vector.<Point>();
stack = new Vector.<MapFilter>();
count = 2;
}
}
}
答案 0 :(得分：1)
您可以使用Floyd Warshall查找每对点之间的最短路径。这将是O(|V|^3)，您不必为每个单元运行它，每次转弯只需运行一次。这是一个如此简单的算法我怀疑它在实践中可能比为每个单元运行BFS / Bellman Ford之类的东西更快。