给定经纬度定位某个城市
方案1
通过GEO HASH，抓取每个格子里的点的地址信息，然后把这个地址信息作为GEO中点坐标的一个信息，当查询给定经纬度是在哪个城市的时候，查询过程是查询距离这个经纬度最近的带有地址的点，认为给定经纬度和最近的这个点是同一个地址下的。通过不断完善GEO中的每个坐标下的地址信息，可以提高整个系统的精度。这种方法相当于一种间接获取的方法的。
方案2
我们知道每个地区是有一定区域的，可以认为是一个闭合的多边形，定位城市，可以理解为判断某个点，是否在某个区域内。
第一点：如何获取城市的区域边界
很幸运的是百度地图提供了这样的API
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>获取地区轮廓线</title>
<script type="text/javascript" src="http://api.map.baidu.com/api?v=1.3">
</script>
<style type="text/css">
body{font-size:13px;margin:10px}
#container{width:800px;height:500px;border:1px solid gray}
</style>
</head>
<body>
<div id="container"></div>
<br />
输入省、直辖市或县名称：<input type="text" id="districtName" style="width:80px" value="重庆市">
<input type="button" onclick="getBoundary()" value="获取轮廓线">
<script type="text/javascript">
var map = new BMap.Map("container");
map.centerAndZoom(new BMap.Point(116.403765, 39.914850), 5);
map.addControl(new BMap.NavigationControl({type: BMAP_NAVIGATION_CONTROL_SMALL}));
map.enableScrollWheelZoom();
function getBoundary(){
var bdary = new BMap.Boundary();
var name = document.getElementById("districtName").value;
bdary.get(name, function(rs){ //获取行政区域
map.clearOverlays(); //清除地图覆盖物
var count = rs.boundaries.length; //行政区域的点有多少个
for(var i = 0; i < count; i++){
var ply = new BMap.Polygon(rs.boundaries[i], {strokeWeight: 2, strokeColor: "#ff0000"}); //建立多边形覆盖物
map.addOverlay(ply); //添加覆盖物
map.setViewport(ply.getPath()); //调整视野
}
});
}
</script>
</body>
</html>
如果没有这样的API，我们也可以变相获得，大多数地图提供了通过经纬度获取地址信息的API，我们可以取一个城市内的一点，然后在城市外周围再取N个点，我们通过二分的方法，可以不断逼近边界
PHP代码
function border($city, $p1, $p2) {
$tpl = 'http://api.map.baidu.com/geocoder/v2/?ak=你的应用key&callback=renderReverse&location=%s,%s&output=json&pois=1';
$i = 5;
while ($i>0) {
$p1X = $p1[0];
$p1Y = $p1[1];
$p2X = $p2[0];
$p2Y = $p2[1];
$dX = abs($p1X - $p2X) / 2;
$dY = abs($p1Y - $p2Y) / 2;
$mX = min($p1X,$p2X) + $dX;
$mY = min($p1Y,$p2Y) + $dY;
$url = sprintf($tpl, $mY, $mX);
$content = rtrim(strstr(file_get_contents($url), "{", false), ")");
$data = json_decode($content, true);
if (($data['result']['addressComponent']['city']) == $city) {
$p1[0] = $mX;
$p1[1] = $mY;
} else {
$p2[0] = $mX;
$p2[1] = $mY;
}
$i--;
}
return array($p1[0],$p1[1]);
}
第二点：知道了区域，如何知道点在区域内
左边的图是从某个点向待检测的点引射线，右边的图是从待检测的点，向某个点引射线，这两种方法都可以，判断射线与闭合区域每条边相交的次数，如果是奇数次的话，可以判断点在区域内，如果是偶数次的话，可以判断点不在区域内。
实际编程中，第二种方法实现起来会更简单一些，而且我们都是向X轴的正方向引射线会更简单。
第三点：如何检测某个点正方向的射线和某条边相交
给定直线P1P2，判断点P平行于X轴正方的射线是否与直线P1P2相交？
首先，满足相交的条件，必须Py夹在P1y和P2y之间。
其次，满足相交的条件，Px小于P1P2最大x，不然肯定不想交了。
最后，P1,P2知道这两个点可以确定一条直线方程y=ax+b，然后p点正方的射线实际是y=Py的直线方程，这两条直线可以求出他们的焦点的x坐标xinters。如果Px小于xinters说明是在直线的y=ax+b的左边，向右引射线肯定能相交，如果Px大于xinters说明在y=ax+b的右边，向右引射线肯定不相交。
附上一段实现的代码
#define MIN(x,y) (x < y ? x : y)
#define MAX(x,y) (x > y ? x : y)
typedef struct {
double x,y;
} Point;
int InsidePolygon(Point *polygon,int N,Point p)
{
int counter = 0;
int i;
double xinters;
Point p1,p2;
p1 = polygon[0];
for (i=1;i<=N;i++) {
p2 = polygon[i % N];
if (p.y > MIN(p1.y,p2.y)) { //低
if (p.y <= MAX(p1.y,p2.y)) { //高
if (p.x <= MAX(p1.x,p2.x)) { //右
if (p1.y != p2.y) { //简单忽略平行X轴这种情况
xinters = (p.y-p1.y)*(p2.x-p1.x)/(p2.y-p1.y)+p1.x; //交叉点坐标 参考./media/point-and-polygon/xinters.jpg
if (p1.x == p2.x || p.x <= xinters)
counter++;
}
}
}
}
p1 = p2;
}
if (counter % 2 == 0)
return 0;
else
return 1;
}
第四：幸运的是JAVA中的awt.Polygon已经实现了类似的算法
以下代码判断了一个经纬度是否在北京范围内
import java.awt.*;
public class Main {
public static void init(Polygon p) {
String s = "117.396814, 40.233095;117.360068, 40.242128;117.348757, 40.25027;117.343034, 40.283717;117.30242, 40.284012;117.300909, 40.298824;117.294971, 40.307248;117.280908, 40.31483;117.277872, 40.331475;117.281586, 40.338248;117.268763, 40.343838;117.248206, 40.376793;117.233105, 40.375358;117.230104, 40.381682;117.235062, 40.392322;117.241661, 40.395752;117.243127, 40.400548;117.246722, 40.40068;117.24604, 40.405045;117.238857, 40.410209;117.240581, 40.42306;117.256385, 40.435421;117.263364, 40.436067;117.26993, 40.447321;117.241196, 40.464298;117.242267, 40.475311;117.232341, 40.484311;117.234676, 40.48752;117.223574, 40.501212;117.215835, 40.502724;117.21653, 40.512643;117.220225, 40.518656;117.239555, 40.519506;117.245336, 40.523751;117.253813, 40.518445;117.266702, 40.525486;117.256848, 40.54366;117.259649, 40.547985;117.256124, 40.554638;117.272602, 40.564079;117.302711, 40.572713;117.317146, 40.582207;117.340905, 40.582418;117.356706, 40.586185;117.372601, 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for ( String t : s.split(";")) {
int x = (int) (Float.parseFloat(t.split(",")[0]) * 1000);
int y = (int) (Float.parseFloat(t.split(",")[1]) * 1000);
p.addPoint(x, y);
}
}
public static boolean contains(Polygon p, float x, float y) {
return p.contains(x * 1000, y * 1000);
}
public static void main(String[] args) {
Polygon p = new Polygon();
init(p);
contains(p, 116.270313f, 39.909622f);
contains(p, 115.590763f, 40.032598f);
contains(p, 115.903517f ,40.875831f);
}
}
