js寻路算法

f()=g()+h();    //g为每个节点到起点得距离;h为每个节点到终点得距离。两个距离通过勾股定理可以算出;

以下代码基本的解释和注释都有。不懂得可以留言哦

css

<style>
        ul{
            list-style: none;
            margin: 0; padding: 0; border: 1px solid #f5f5f5; border-right: none; border-bottom: none; margin: 100px auto; } ul li{ float: left; border: solid 1px #f5f5f5; border-left: none; border-top: none; display: inline-block; background: black; } .start{ background: skyblue; } .end{ background: green; } .usual{ background: pink; } </style>

 html

<body>
    <ul id="map-ul"></ul> <input id="btn" type="button" value="开始" /> </body>

js

<script>
    var oUl = document.getElementById("map-ul");
    var startBtn = document.getElementById("btn");
    var Li = oUl.getElementsByTagName("li"); var beginLi = document.getElementsByClassName("start"); var endLi = document.getElementsByClassName("end"); var map = [ 1,1,1,1,1,1,1,1,1,1,//0为起点;2为障碍物,3为终点 1,1,0,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1,1,1, 1,1,1,1,1,2,2,2,1,1, 1,1,1,1,1,2,1,1,1,1, 1,1,1,1,1,2,1,1,1,1, 1,1,1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1,1,3, 1,1,1,1,1,1,1,1,1,1 ]; var openArr = [];//可能要走的路线 var closeArr = [];//已经关闭的路线 var cols = Math.sqrt(map.length) var sizeBox = 20; //初始化  init() function init(){ creatMap();//初始化地图 startBtn.onclick = function(){ openFn(); } } function creatMap(){ oUl.style.width = cols * (sizeBox + 1) + 1 + "px"; oUl.style.height = cols * (sizeBox + 1) + 1 + "px"; for (var i=0;i<map.length;i++) { var oLi = document.createElement("li"); oLi.style.width = sizeBox + "px"; oLi.style.height = sizeBox + "px"; oUl.appendChild(oLi) if (map[i] == 2) { oLi.className = "usual" closeArr.push(oLi)//障碍物添加到关闭得线路 } else if(map[i] == 0){ oLi.className = "start" openArr.push(oLi)//起始点添加到要走得线路 }else if(map[i] == 3){ oLi.className = "end" } } } function f(nodeLi){//nodeLi每个li节点 return g(nodeLi) + h(nodeLi);//g为节点到初始位置的距离,h为节点到结束位置的距离  } function g(nodeLi){ //勾股定理算出距离 var x = beginLi[0].offsetLeft - nodeLi.offsetLeft; var y = beginLi[0].offsetTop - nodeLi.offsetTop; return Math.sqrt(x*x+y*y); } function h(nodeLi){ var x = endLi[0].offsetLeft - nodeLi.offsetLeft; var y = endLi[0].offsetTop - nodeLi.offsetTop; return Math.sqrt(x*x+y*y); } function openFn(){ //以起始位置为可能要走的第一个Li var nowLi = openArr.shift(); if(nowLi == endLi[0]){ showLine(); return; } closeFn(nowLi);//将找过得节点添加到关闭的路线 findLi(nowLi);//寻找第一个Li周围的可能要走的Li openArr.sort(function(li1,li2){//将可能走得线路得节点通过距离从下到大排序。当下次循环得时候可以直接拿到上个最近得节点, return li1.num - li2.num; }) openFn(); } function findLi(nowLi){ var result = [];//所有没走过的路线。去除closeArr里存在的路线 for (var i=0;i<Li.length;i++) { if( filter(Li[i])){ result.push(Li[i]) } } function filter(Li){//如果closeArr里没有就添加到队列 for (var i=0;i<closeArr.length;i++) { if(closeArr[i] == Li){ return false; } } for (var i=0;i<openArr.length;i++) { if(openArr[i] == Li){ return false; } } return true; } //在所有没走过的节点找到与当前节点相邻的8个方向的节点 for(var i=0;i<result.length;i++){ if(Math.abs(nowLi.offsetLeft - result[i].offsetLeft)<=21&&Math.abs(nowLi.offsetTop - result[i].offsetTop)<=21){ result[i].num = f(result[i]); result[i].parent = nowLi;//设置指针  openArr.push(result[i]) } } } function showLine(){ var result = []; var lastLi = closeArr.pop(); var iNow = 0; findParent(lastLi); function findParent(li){ result.unshift(li); if( li.parent == beginLi[0] ){ return; } findParent(li.parent);//通过指针递归  console.log(li.parent) } for (var i=0;i<result.length;i++) { result[i].style.backgroundColor = "plum"; } } function closeFn(nowLi){ closeArr.push(nowLi); } </script>

    

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转载自www.cnblogs.com/fangling/p/9210906.html