Treasure Hunt--POJ 1066
1、题目类型：计算几何、线段相交。
2、解题思路：分析，读题后第一想法是从treasure点多边形向外做BFS直到到达边界，但区域内的各个不规则多边形无法确定；后发现只要在区域的四周的焦点间中点做与treasure点的线段，获取其最小焦点就可（即使线段通过内部线段的交点此时不影响结果，同样计算了两次）。步骤，（1）对区域四个边界上的点进行排序（源码的方法比较呆板、代码比较冗余，应该有更好的方法）；（2）对四个边界上，相邻焦点的中点与treasure点做线段并与其他线段判断相交，获得最小交点树的情况，即为此题最优。
3、注意事项：注意计算几何模板的应用，判断线段相交的点是有序的。
4、实现方法:
#include
<
iostream
>
#include
<
algorithm
>
using
namespace
std;
const
double
eps
=
1e
-
10
;
struct
point
{
double
x, y;
};
struct
Tline
{
point p1,p2;
};
point side[
4
][
35
],End;
Tline line[
31
];
int
num[
4
],ans,n;
double
min(
double
a,
double
b) {
return
a
<
b
?
a : b; }
double
max(
double
a,
double
b) {
return
a
>
b
?
a : b; }
bool
inter(point a, point b, point c, point d)
{
if
( min(a.x, b.x)
>
max(c.x, d.x)
||
min(a.y, b.y)
>
max(c.y, d.y)
||
min(c.x, d.x)
>
max(a.x, b.x)
||
min(c.y, d.y)
>
max(a.y, b.y) )
return
0
;
double
h, i, j, k;
h
=
(b.x
-
a.x)
*
(c.y
-
a.y)
-
(b.y
-
a.y)
*
(c.x
-
a.x);
i
=
(b.x
-
a.x)
*
(d.y
-
a.y)
-
(b.y
-
a.y)
*
(d.x
-
a.x);
j
=
(d.x
-
c.x)
*
(a.y
-
c.y)
-
(d.y
-
c.y)
*
(a.x
-
c.x);
k
=
(d.x
-
c.x)
*
(b.y
-
c.y)
-
(d.y
-
c.y)
*
(b.x
-
c.x);
return
h
*
i
<=
eps
&&
j
*
k
<=
eps;
}
bool
cmp(point p1,point p2)
{
return
(p1.y
==
p2.y
&&
p1.x
<
p2.x)
||
(p1.x
==
p2.x
&&
p1.y
<
p2.y);
}
void
Solve()
{
int
i,j;
memset(num,
0
,
sizeof
(num));
for
(i
=
0
;i
<
n;i
++
)
{
if
(line[i].p1.x
==
0
){
side[
0
][num[
0
]].x
=
0
;
side[
0
][num[
0
]].y
=
line[i].p1.y;
num[
0
]
++
;}
if
(line[i].p2.x
==
0
){
side[
0
][num[
0
]].x
=
0
;
side[
0
][num[
0
]].y
=
line[i].p2.y;
num[
0
]
++
;}
if
(line[i].p1.x
==
100
){
side[
1
][num[
1
]].x
=
100
;
side[
1
][num[
1
]].y
=
line[i].p1.y;
num[
1
]
++
;}
if
(line[i].p2.x
==
100
){
side[
1
][num[
1
]].x
=
100
;
side[
1
][num[
1
]].y
=
line[i].p2.y;
num[
1
]
++
;}
if
(line[i].p1.y
==
0
){
side[
2
][num[
2
]].x
=
line[i].p1.x;
side[
2
][num[
2
]].y
=
0
;
num[
2
]
++
;}
if
(line[i].p2.y
==
0
){
side[
2
][num[
2
]].x
=
line[i].p2.x;
side[
2
][num[
2
]].y
=
0
;
num[
2
]
++
;}
if
(line[i].p1.y
==
100
){
side[
3
][num[
3
]].x
=
line[i].p1.x;
side[
3
][num[
3
]].y
=
100
;
num[
3
]
++
;}
if
(line[i].p2.y
==
100
){
side[
3
][num[
3
]].x
=
line[i].p2.x;
side[
3
][num[
3
]].y
=
100
;
num[
3
]
++
;}
}
side[
0
][num[
0
]].x
=
0
;
side[
0
][num[
0
]].y
=
100
;
side[
1
][num[
1
]].x
=
100
;
side[
1
][num[
1
]].y
=
100
;
side[
2
][num[
2
]].x
=
100
;
side[
2
][num[
2
]].y
=
0
;
side[
3
][num[
3
]].x
=
100
;
side[
3
][num[
3
]].y
=
100
;
for
(i
=
0
;i
<
4
;i
++
)
sort(side[i],side[i]
+
num[i],cmp);
point p,q,mid;
int
m,k;
for
(i
=
0
;i
<
4
;i
++
)
{
p.x
=
0
;
p.y
=
0
;
for
(j
=
0
;j
<=
num[i];j
++
)
{
m
=
1
;
q
=
side[i][j];
mid.x
=
(q.x
+
p.x)
*
0.5
;
mid.y
=
(q.y
+
p.y)
*
0.5
;
for
(k
=
0
;k
<
n;k
++
)
if
(inter(mid,End,line[k].p1,line[k].p2))
m
++
;
if
(m
<
ans)
ans
=
m;
p
=
q;
}
}
}
int
main()
{
int
i;
memset(side,
0
,
sizeof
(side));
scanf(
"
%d
"
,
&
n);
for
(i
=
0
;i
<
n;i
++
)
{
scanf(
"
%lf%lf%lf%lf
"
,
&
line[i].p1.x,
&
line[i].p1.y,
&
line[i].p2.x,
&
line[i].p2.y);
}
scanf(
"
%lf%lf
"
,
&
End.x,
&
End.y);
ans
=
99999
;
Solve();
cout
<<
"
Number of doors =
"
<<
ans
<<
endl;
return
0
;
}