A Round Peg in a Ground Hole--POJ 1584
1、题目类型：模拟、计算几何。
2、解题思路：（1）检测是否为凹多边形；（2）检测圆心是否在凸多边形内部；（3）检测圆心到多边形所有边的距离是否小于半径 （具体见实现代码）。
3、注意事项：注意多边形的顺时针和逆时针，由于上的是模板，所以细节考虑并不多。
4、实现方法:
#include
<
iostream
>
#include
<
cmath
>
#define
MIN(x,y) (x < y ? x : y)
#define
MAX(x,y) (x > y ? x : y)
using
namespace
std;
struct
Point
{
double
x,y;
};
int
n;
double
r;
Point P[
160
];
Point c;
//
检测是否为凹多边形: *p表示为多边形序列，n表示为多边形顶点数
int
Check1(Point
*
p,
int
n)
{
int
i,j,k,flag
=
0
;
double
z;
if
(n
<
3
)
return
(
0
);
for
(i
=
0
;i
<
n;i
++
)
{
j
=
(i
+
1
)
%
n;
k
=
(i
+
2
)
%
n;
z
=
(p[j].x
-
p[i].x)
*
(p[k].y
-
p[j].y);
z
-=
(p[j].y
-
p[i].y)
*
(p[k].x
-
p[j].x);
if
(z
<
0
)
flag
|=
1
;
else
if
(z
>
0
)
flag
|=
2
;
if
(flag
==
3
)
return
0
;
}
if
(flag
!=
0
)
return
1
;
else
return
0
;
}
//
检测点是否在多边形内: *polygon表示为多边形序列，n表示为多边形顶点数,p表示为点
int
Check2(Point
*
polygon,
int
N,Point p)
{
int
i,counter
=
0
;
double
xinters;
Point p1,p2;
p1
=
polygon[
0
];
for
(i
=
1
;i
<=
N;i
++
)
{
p2
=
polygon[i
%
N];
if
(p.y
>
MIN(p1.y,p2.y))
{
if
(p.y
<=
MAX(p1.y,p2.y))
{
if
(p.x
<=
MAX(p1.x,p2.x))
{
if
(p1.y
!=
p2.y)
{
xinters
=
(p.y
-
p1.y)
*
(p2.x
-
p1.x)
/
(p2.y
-
p1.y)
+
p1.x;
if
(p1.x
==
p2.x
||
p.x
<=
xinters)
counter
++
;
}
}
}
}
p1
=
p2;
}
if
(counter
%
2
==
0
)
return
1
;
else
return
0
;
}
//
计算点到线段的最短距离：p1,p2表示线段的两点，q表示距离点
double
Mindistance(Point p1,Point p2,Point q)
{
int
flag
=
1
;
double
k;
Point s;
if
(p1.x
==
p2.x) {s.x
=
p1.x;s.y
=
q.y;flag
=
0
;}
if
(p1.y
==
p2.y) {s.x
=
q.x;s.y
=
p1.y;flag
=
0
;}
if
(flag)
{
k
=
(p2.y
-
p1.y)
/
(p2.x
-
p1.x);
s.x
=
(k
*
k
*
p1.x
+
k
*
(q.y
-
p1.y)
+
q.x)
/
(k
*
k
+
1
);
s.y
=
k
*
(s.x
-
p1.x)
+
p1.y;
}
if
(MIN(p1.x,p2.x)
<=
s.x
&&
s.x
<=
MAX(p1.x,p2.x))
return
sqrt((q.x
-
s.x)
*
(q.x
-
s.x)
+
(q.y
-
s.y)
*
(q.y
-
s.y));
else
return
MIN(sqrt((q.x
-
p1.x)
*
(q.x
-
p1.x)
+
(q.y
-
p1.y)
*
(q.y
-
p1.y)),
sqrt((q.x
-
p2.x)
*
(q.x
-
p2.x)
+
(q.y
-
p2.y)
*
(q.y
-
p2.y)));
}
int
Check3()
{
int
i;
P[n].x
=
P[
0
].x;
P[n].y
=
P[
0
].y;
for
(i
=
0
;i
<
n;i
++
)
{
if
(Mindistance(P[i],P[i
+
1
],c)
<
r)
return
0
;
}
return
1
;
}
void
Solve()
{
if
(
!
Check1(P,n))
{
cout
<<
"
HOLE IS ILL-FORMED
"
<<
endl;
return
;
}
if
(Check2(P,n,c))
{
cout
<<
"
PEG WILL NOT FIT
"
<<
endl;
return
;
}
if
(
!
Check3())
{
cout
<<
"
PEG WILL NOT FIT
"
<<
endl;
return
;
}
else
{
cout
<<
"
PEG WILL FIT
"
<<
endl;
return
;
}
}
int
main()
{
int
i;
while
(cin
>>
n
&&
n
>=
3
)
{
cin
>>
r
>>
c.x
>>
c.y;
for
(i
=
0
;i
<
n;i
++
)
{
cin
>>
P[i].x
>>
P[i].y;
}
Solve();
}
return
0
;
}