题解：
投在地上的影子是很多圆和两圆公切线组成的梯形的面积并。
PS：圆只要和地面平行，无论光从哪个角度照射，投影都是圆
其实应该一开始应先分成若干份做simpson的。。
View Code
1
#include <iostream>
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#include <cstring>
3
#include <cstdio>
4
#include <cstdlib>
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#include <algorithm>
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#include <cmath>
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#define
N 520
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#define
EPS 1e-6
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using
namespace
std;
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int
n;
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double
a[N],b[N],af,h,lt,rt;
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struct
C
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{
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double
x,y,p,q;
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}c[N];
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inline
int
dc(
double
x)
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{
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if
(x>EPS)
return
1
;
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else
if
(x<-EPS)
return
-
1
;
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return
0
;
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}
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inline
void
read()
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{
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scanf(
"
%d%lf
"
,&n,&
af);
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af=
1
/
tan(af);
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for
(
int
i=
1
;i<=n+
1
;i++
)
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{
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scanf(
"
%lf
"
,&
a[i]);
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h+=a[i]; a[i]=h*
af;
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}
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lt=a[
1
];rt=a[n+
1
];
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for
(
int
i=
1
;i<=n;i++
)
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{
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scanf(
"
%lf
"
,&
b[i]);
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lt=min(lt,a[i]-
b[i]);
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rt=max(rt,a[i]+
b[i]);
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}
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}
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inline
void
calc()
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{
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for
(
int
i=
1
;i<=n;i++
)
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if
(a[i+
1
]-a[i]>fabs(b[i+
1
]-
b[i]))
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{
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c[i].x=a[i]+b[i]*(b[i]-b[i+
1
])/(a[i+
1
]-
a[i]);
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c[i].y=sqrt(b[i]*b[i]-(c[i].x-a[i])*(c[i].x-
a[i]));
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c[i].p=a[i+
1
]+b[i+
1
]*(b[i]-b[i+
1
])/(a[i+
1
]-
a[i]);
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c[i].q=sqrt(b[i+
1
]*b[i+
1
]-(c[i].p-a[i+
1
])*(c[i].p-a[i+
1
]));
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}
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}
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inline
double
f(
double
p)
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{
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double
res=
0
;
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for
(
int
i=
1
;i<=n;i++
)
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{
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if
(fabs(a[i]-p)<b[i]) res=max(res,sqrt(b[i]*b[i]-(a[i]-p)*(a[i]-
p)));
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if
(p>c[i].x&&p<c[i].p) res=max(res,c[i].y-(p-c[i].x)*(c[i].y-c[i].q)/(c[i].p-
c[i].x));
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}
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return
res;
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}
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inline
double
simpson(
double
l,
double
r,
double
fl,
double
fmid,
double
fr)
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{
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return
(fl+fr+
4
*fmid)*(r-l)/
6
;
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}
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inline
double
rsimpson(
double
l,
double
r,
double
fl,
double
fmid,
double
fr)
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{
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double
p,q,mid,x,y,z;
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mid=(l+r)/
2
;
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p=f((l+mid)/
2
); q=f((mid+r)/
2
);
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x=simpson(l,r,fl,fmid,fr); y=simpson(l,mid,fl,p,fmid); z=
simpson(mid,r,fmid,q,fr);
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if
(dc(fabs(x-y-z))==
0
)
return
y+
z;
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else
return
rsimpson(l,mid,fl,p,fmid)+
rsimpson(mid,r,fmid,q,fr);
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}
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inline
void
go()
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{
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calc();
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printf(
"
%.2lf\n
"
,
2
*rsimpson(lt,rt,
0
,f(lt+rt)/
2
,
0
));
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}
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int
main()
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{
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read(),go();
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return
0
;
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}
