我想查看一个点是否在多边形内,当然我已经过谷歌搜索并查看之前是否有人回答这个问题,并找到了这个算法:http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html。
这个算法很好用,除非多边形是部分开放的。例如:
A-E被检测到了,但B多边形的开放部分也被认为是封闭的!如果你运行这个示例代码,你就会知道我的意思。
#include <stdio.h>
int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
int i, j, c = 0;
for (i = 0, j = nvert-1; i < nvert; j = i++) {
if ( ((verty[i]>testy) != (verty[j]>testy)) &&
(testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
c = !c;
}
return c;
}
int main(int argc, char *argv[])
{
// 1 closed [A]
float x1[] = { 0, 1, 1, 0, 0 };
float y1[] = { 0, 0, 1, 1, 0 };
printf("1: %d (1 expected)\n", pnpoly(5, x1, y1, 0.8, 0.8));
printf("1: %d (0 expected)\n", pnpoly(5, x1, y1, -0.8, -0.8));
// 1 closed [B] with a partial open
// please note that the vertex between [0,-1] and [0,0] is missing
float x2[] = { 0, 1, 1, 0, 0, -1, -1, 0 };
float y2[] = { 0, 0, 1, 1, 0, 0, -1, -1 };
printf("2: %d (1 expected)\n", pnpoly(8, x2, y2, 0.8, 0.8));
printf("2: %d (0 expected)\n", pnpoly(8, x2, y2, -0.8, -0.8)); // <- fails
// 2 closed [C/D/E]
float x3[] = { 0, 1, 1, 0, 0, -1, -1, 0, 0 };
float y3[] = { 0, 0, 1, 1, 0, 0, -1, -1, 0 };
printf("3: %d (1 expected)\n", pnpoly(9, x3, y3, 0.8, 0.8));
printf("3: %d (1 expected)\n", pnpoly(9, x3, y3, -0.8, -0.8));
return 0;
}
x2/y2多边形由一个封闭块连接到一个部分开放的块组成。 pnpoly函数仍然认为在开放块内的点在多边形内。
我的问题是:我该如何解决这个问题?或者我有所忽略吗?
提前感谢。