javascript多边形碰撞检测详解编程语言

 const app = new PIXI.Application({ antialias: true }); 

document.body.appendChild(app.view); 

const graphics = new PIXI.Graphics(); 


 // the reason for this is because of multitouch 

 // we want to track the movement of this particular touch 

 this.data = event.data; 

 this.alpha = 0.5; 

 this.dragging = true; 

function onDragEnd() { 

 this.alpha = 1; 

 this.dragging = false; 

 // set the interaction data to null 

 this.data = null; 

 var posPolygon=[]; 

 var dianlist={}; 

 dianlist[x]=600; 

 dianlist[y]=370; 

 posPolygon.push(dianlist) 

 var dianlist={}; 

 dianlist[x]=700; 

 dianlist[y]=460; 

 posPolygon.push(dianlist) 

 var dianlist={}; 

 dianlist[x]=780; 

 dianlist[y]=420; 

 posPolygon.push(dianlist) 

 var dianlist={}; 

 dianlist[x]=730; 

 dianlist[y]=570; 

 posPolygon.push(dianlist) 

 var dianlist={}; 

 dianlist[x]=590; 

 dianlist[y]=520; 

 posPolygon.push(dianlist) 

function onDragMove() { 

 if (this.dragging) { 

 const newPosition = this.data.getLocalPosition(this.parent); 

 this.x = newPosition.x; 

 this.y = newPosition.y; 

 var baoweihe=this.getBounds(); 

 var youxiajiaox=baoweihe.x+baoweihe.width; 

 var youxiajiaoy=baoweihe.y+baoweihe.height; 

 var poslist=[]; 

 var pos={}; 

 pos[x]=baoweihe.x; 

 pos[y]=baoweihe.y; 

 poslist.push(pos); 

 var pos={}; 

 pos[x]=youxiajiaox; 

 pos[y]=baoweihe.y; 

 poslist.push(pos); 

 var pos={}; 

 pos[x]=youxiajiaox; 

 pos[y]=youxiajiaoy; 

 poslist.push(pos); 

 var pos={}; 

 pos[x]=baoweihe.x; 

 pos[y]=youxiajiaoy; 

 poslist.push(pos); 


function PolygonInPolygon(posPolygonA, posPolygonB, count){ 

 console.log(posPolygonA); 

 var count1=posPolygonA.length; 

 for(var i=0;i count1;i++ ){ 

 var pos=posPolygonA[i]; 

 console.log(pos); 

 var ispengzhuang=PointInPolygon( pos, posPolygonB, count); 

 if(ispengzhuang){ 

 alert(碰撞了) 

function PointInPolygon( pos, posPolygonB, count) 

 var cross = 0; //交点个数 

 for( var i = 0; i count; i++ ) 

 var p1 = posPolygon[i]; 

 var p2 = posPolygon[(i + 1) % count]; //下一个节点 

 // p1p2这条边与水平线平行 

 if( p1.y == p2.y ) 

 continue; 

 // 交点在p1p2的延长线上 

 if( pos.y Math.min( p1.y, p2.y ) ) 

 continue; 

 // 交点在p1p2的延长线上 

 if( pos.y Math.max( p1.y, p2.y ) ) 

 continue; 

 // 计算交点 X 左边 ： (p2.y - p1.y)/(p2.x - p1.x) = (y - p1.y)/(x - p1.x) 

 // 直线 K 值相等， 交点y = pos.y 

 let x = (pos.y - p1.y) * (p2.x - p1.x) / (p2.y - p1.y) + p1.x 

 // 只统计单边交点，即点的正向方向 

 if(x pos.x) 

 cross ++; 

 return cross % 2 == 1; 

}

原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/18770.html