方法1：過該點作一條射線，然后判斷該射線與三條邊的相交情況；
方法2：分別連接P點和三個頂點A,B,C，然后判斷矢量PA,PB,PC所夾角之和是否為360度，如果是360度則P在三角形內
1.
#include <stdio.h>
#include <math.h>
struct TPoint
{
float x;
float y;
};
float area(struct TPoint p1,struct TPoint p2,struct TPoint p3){
return fabs((p1.x-p3.x)*(p2.y-p3.y)-(p2.x-p3.x)*(p1.y-p3.y));
}
float mul(struct TPoint p1,struct TPoint p2,struct TPoint p0){
return((p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y));
}
int inside(struct TPoint tr[],struct TPoint p){
int i;
for (i=0;i<3;i++)
if (mul(p,tr[i],tr[(i+1)%3])*mul(p,tr[(i+2)%3],tr[(i+1)%3])>0)
return 0;
return 1;
}
int inside2(struct TPoint tr[],struct TPoint p){
if (fabs(area(tr[0],tr[1],tr[2])-
area(p,tr[1],tr[2])-
area(tr[0],p,tr[2])-
area(tr[0],tr[1],p))<1.0e-20)
return 1;
else
return 0;
}
main(){
struct TPoint tr[3]={{-1,1},{1,0},{3,0}},p={1,2};
printf("algorithm 1:");
if (inside(tr,p))
printf("In\n");
else
printf("Out\n");
printf("algorithm 2:");
if (inside2(tr,p))
printf("In\n");
else
printf("Out\n");
}
2
就是判斷點到三個頂點的夾角之和為2*PI
// 返回1 表示在那，0表示在外
int inside3(const struct TPoint tr[], struct TPoint p)
{
TPoint p0,p1,p2;
p0.x = tr[0].x - p.x ; p0.y = tr[0].y - p.y;
p1.x = tr[1].x - p.x ; p1.y = tr[1].y - p.y;
p2.x = tr[2].x - p.x ; p2.y = tr[2].y - p.y;
double arg1 = acos((p0.x*p1.x + p0.y*p1.y)/sqrt(p0.x*p0.x + p0.y*p0.y)/sqrt(p1.x*p1.x+p1.y*p1.y));
double arg2 = acos((p0.x*p2.x + p0.y*p2.y)/sqrt(p0.x*p0.x + p0.y*p0.y)/sqrt(p2.x*p2.x+p2.y*p2.y));
double arg3 = acos((p2.x*p1.x + p2.y*p1.y)/sqrt(p2.x*p2.x + p2.y*p2.y)/sqrt(p1.x*p1.x+p1.y*p1.y));
if( fabs(2*3.1415926-arg1-arg2-arg3)<0.0001 ) return 1;
return 0;
}