將圓圈組合爲多邊形而沒有與公共區域相交

Question

我正在爲我公司的供應商創建覆蓋圖，然後需要檢查我們的客戶中有多少人生活在整體覆蓋範圍之外。供應商的覆蓋區域始終是一個圓圈。

我使用this解決方案將我們供應商的覆蓋區域合併爲一個多邊形，以便我可以使用containsLocation function來比較客戶的位置。問題是containsLocation函數將一些重疊的覆蓋區域標識爲在覆蓋範圍之外。以下是一個示例，其中該引腳應該保持隱形，因爲它位於2個供應商的覆蓋範圍內。 Overlapping Vendor Coverage Example

function updateMap(){ 
    var radius = $('#radius').val(); 
    $.ajax({ 
     url: "", 
     dataType: "json", 
     type: "POST", 
     data: $('#dataForm').serialize() 
    }).done(function(result){ 
     for(var i=0; i<result.length; i++){ 
     center = {lat: parseFloat(result[i].Latitude), lng: parseFloat(result[i].Longitude)}; 
     shapeArray.push(drawCircle(center, (radius), 1)); 
     } 
     coverage = new google.maps.Polygon({ 
      paths: shapeArray, 
      strokeColor: "#ff0000", 
      strokeOpacity: 0.8, 
      strokeWeight: 1, 
      fillColor: "#ff0000", 
      fillOpacity: 0.35, 
      map: map 
     }); 
    }); 
    } 

    function drawCircle(point, radius, dir) 
    { 
    var d2r = Math.PI/180; // degrees to radians 
    var r2d = 180/Math.PI; // radians to degrees 
    var earthsradius = 3963; // 3963 is the radius of the earth in miles 
    var points = 32; 

    // find the raidus in lat/lon 
    var rlat = (radius/earthsradius) * r2d; 
    var rlng = rlat/Math.cos(point.lat * d2r); 

    var extp = new Array(); 
    if (dir==1) {var start=0;var end=points+1} // one extra here makes sure we connect the 
    else{var start=points+1;var end=0} 
    for (var i=start; (dir==1 ? i < end : i > end); i=i+dir) 
    { 
     var theta = Math.PI * (i/(points/2)); 
     ey = point.lng + (rlng * Math.cos(theta)); // center a + radius x * cos(theta) 
     ex = point.lat + (rlat * Math.sin(theta)); // center b + radius y * sin(theta) 
     extp.push(new google.maps.LatLng(ex, ey)); 
    } 
    return extp; 
    } 

    function showUncovered(el){ 
    if(!el.checked){ 
     for(var i=0; i<markerArray.length; i++){ 
      markerArray[i].setVisible(false); 
     } 
     return; 
    } 
    for(var i=0;i<markerArray.length;i++){ 
     var pos = markerArray[i].position; 
     var isAffected = false; 
     if(!google.maps.geometry.poly.containsLocation(pos,coverage)){ 
     markerArray[i].setVisible(true); 
     }  
    }   
    }  

Answer 1

谷歌接受了這個bug。 https://code.google.com/p/gmaps-api-issues/issues/detail?id=10609