PKU1039 Pipe 计算几何线段交点
模板：给定L1上两点（ax,ay），（bx,by），L2上两点（cx,cy），（dx,dy），求两直线交点（x,y）。
double
cross(
double
x1,
double
y1,
double
x2,
double
y2)
{
return
x1
*
y2
-
x2
*
y1;
}
void
intersec(point a, point b, point c, point d,
double
*
x,
double
*
y)
{
double
u
=
cross(d.x
-
a.x, d.y
-
a.y, b.x
-
a.x, b.y
-
a.y);
double
v
=
cross(c.x
-
a.x, c.y
-
a.y, b.x
-
a.x, b.y
-
a.y);
double
w
=
u
-
v;
*
x
=
(u
*
1.0
*
c.x
-
v
*
1.0
*
d.x)
/
(w
*
1.0
);
*
y
=
(u
*
1.0
*
c.y
-
v
*
1.0
*
d.y)
/
(w
*
1.0
);
}
| 1039 | Accepted | 200K | 63MS |
63MS是有些慢，这一题是刘汝佳的《算法艺术与信息学竞赛》P359的例题。
思路：枚举上下两个顶点成光线所在直线，然后判断光线是否合法，合法的话枚举判断光线与管道上下壁是否相交，并存下最远的交点横坐标。
代码
#include
<
iostream
>
#include
<
cmath
>
using
namespace
std;
#define
MAXN 25
#define
eps 1e-8
struct
point{
double
x,y;
point(
double
xx
=
0
,
double
yy
=
0
){
x
=
xx; y
=
yy;
}
point
operator
-
(
const
point
&
a)
const
{
return
point(x
-
a.x, y
-
a.y);
}
double
operator
*
(
const
point
&
a)
const
{
//
叉积
return
x
*
a.y
-
y
*
a.x;
}
void
input() {
scanf(
"
%lf%lf
"
,
&
x,
&
y);
}
}upper[MAXN],botto[MAXN];
int
n;
int
sgn(
double
x)
{
return
(x
>
eps)
-
(x
<-
eps);
}
void
intersec(point a, point b, point c, point d,
double
*
x,
double
*
y)
{
double
u
=
(d
-
a)
*
(b
-
a);
double
v
=
(c
-
a)
*
(b
-
a);
double
w
=
u
-
v;
*
x
=
(u
*
c.x
-
v
*
d.x)
/
w;
*
y
=
(u
*
c.y
-
v
*
d.y)
/
w;
}
//
判断直线L（a,b）和线段L（up,down）是否相交
bool
check(point a,point b,point up,point down)
{
double
tx,ty;
intersec(a,b,up,down,
&
tx,
&
ty);
if
(sgn(ty
-
up.y)
<=
0
&&
sgn(ty
-
down.y)
>=
0
)
return
true
;
return
false
;
}
void
slove()
{
int
i,j,k,flag;
double
maxx,tx,ty;
maxx
=
upper[
1
].x;
//
初始化
for
(i
=
1
;i
<=
n;i
++
)
{
for
(j
=
1
;j
<=
n;j
++
)
{
flag
=
0
;
if
(i
==
j)
continue
;
for
(k
=
(i
>
j
?
i:j);k
>=
0
;k
--
)
{
if
(check(upper[i],botto[j],upper[k],botto[k])
==
0
)
break
;
}
if
(k
>=
1
)
continue
;
for
(k
=
1
;k
<=
n;k
++
)
{
if
(check(upper[i],botto[j],upper[k],botto[k])
==
0
)
break
;
}
if
(k
>
n) { flag
=
1
;
break
;}
intersec(upper[i],botto[j],upper[k
-
1
],upper[k],
&
tx,
&
ty);
if
(sgn(tx
-
upper[k
-
1
].x)
>=
0
&&
sgn(tx
-
upper[k].x)
<=
0
)
if
(maxx
<
tx) maxx
=
tx;
intersec(upper[i],botto[j],botto[k
-
1
],botto[k],
&
tx,
&
ty);
if
(sgn(tx
-
botto[k
-
1
].x)
>=
0
&&
sgn(tx
-
botto[k].x)
<=
0
)
if
(maxx
<
tx) maxx
=
tx;
}
if
(flag)
break
;
}
if
(flag) puts(
"
Through all the pipe.
"
);
else
printf(
"
%.2lf\n
"
,maxx);
}
int
main()
{
while
(scanf(
"
%d
"
,
&
n),n)
{
for
(
int
i
=
1
;i
<=
n;i
++
)
{
upper[i].input();
botto[i].x
=
upper[i].x;
botto[i].y
=
upper[i].y
-
1
;
}
slove();
}
return
0
;
}