HDU2912 Spherical Mirrors 计算几何
题意：求一条射线经过一系列球反射后的反射点
分析：从起点出发，求出射线与球的最近交点，然后更改反射线为入射线循环找射线与球的最近交点，直到没有球与射线相交。
#include
<
iostream
>
#include
<
cmath
>
using
namespace
std;
struct
point{
double
x,y,z;
point(
double
xx
=
0
,
double
yy
=
0
,
double
zz
=
0
){
//
构造函数
x
=
xx,y
=
yy,z
=
zz;
}
point
operator
^
(
double
h){
return
point(x
*
h,y
*
h,z
*
h);
}
double
operator
^
(point h){
//
点乘
return
x
*
h.x
+
y
*
h.y
+
z
*
h.z;
}
point
operator
+
(point h){
return
point(x
+
h.x,y
+
h.y,z
+
h.z);
}
point
operator
-
(point h){
return
point(x
-
h.x,y
-
h.y,z
-
h.z);
}
point
operator
*
(point b){
//
叉乘
return
point(y
*
b.z
-
b.y
*
z, z
*
b.x
-
b.z
*
x, x
*
b.y
-
b.x
*
y);
}
double
len2()
const
{
return
x
*
x
+
y
*
y
+
z
*
z;
}
double
len()
const
{
return
sqrt(len2());
}
point turnlen(
double
l)
const
{
//
改变长度
double
r
=
l
/
len();
return
point(x
*
r,y
*
r,z
*
r);
}
void
input()
{
scanf(
"
%lf%lf%lf
"
,
&
x,
&
y,
&
z);
}
}p1,p2,ans;
struct
sphere{
point c;
double
r;
void
input(){
c.input();
scanf(
"
%lf
"
,
&
r);
}
}sph[
110
];
int
n;
const
double
eps
=
1e
-
8
;
int
sgn(
double
x)
{
if
(fabs(x)
<
eps)
return
0
;
return
x
>
0
?
1
:
-
1
;
}
bool
intersection(sphere s,point p1,point p2,point
&
c)
//
求p2-p1与球s的交点c
{
double
d
=
((p1
-
s.c)
*
(p2
-
s.c)).len()
/
(p1
-
p2).len();
//
球心到p2-p1的距离
if
(sgn(d
-
s.r)
>=
0
)
return
false
;
//pp 球心与直线p2-p1的垂足
point pp
=
(sgn(d)
==
0
?
s.c:s.c
+
((p1
-
s.c)
*
(p2
-
s.c)
*
(p1
-
p2)).turnlen(d) );
if
(sgn((pp
-
p1)
^
(p2
-
p1))
<=
0
)
//
如果pp在p2-p1的反向延长线上 返回false
return
false
;
c
=
pp
+
(p1
-
p2).turnlen(sqrt(s.r
*
s.r
-
d
*
d));
return
true
;
}
point reflection(point p1,point rep,point c)
//
p1关于rep-c的对称点
{
double
d
=
((rep
-
p1)
*
(c
-
p1)).len()
/
(rep
-
c).len();
return
p1
+
((rep
-
p1)
*
(c
-
p1)
*
(rep
-
c)).turnlen(d
*
2.0
);
}
void
compute()
{
while
(
true
){
point c;
int
k
=-
1
;
for
(
int
i
=
0
;i
<
n;i
++
)
//
找到与射线相交的最近的球
{
point tmp;
if
(intersection(sph[i],p1,p2,tmp)){
if
(k
==-
1
||
sgn((tmp
-
p1).len()
-
(c
-
p1).len())
<
0
){
c
=
tmp;
k
=
i;
}
}
}
if
(k
==-
1
)
break
;
p2
=
reflection(p1,sph[k].c,c);
//
p2反射线上一点
p1
=
c;
//
反射点
}
ans
=
p1;
}
int
main()
{
while
(scanf(
"
%d
"
,
&
n),n)
{
p1
=
point();
p2.input();
for
(
int
i
=
0
;i
<
n;i
++
)
sph[i].input();
compute();
printf(
"
%.8lf %.8lf %.8lf\n
"
,ans.x,ans.y,ans.z);
}
return
0
;
}