Fishnet--POJ 1408
1、题目类型：计算几何。
2、解题思路：（1）建立点结构体Node；（2）给四条边上的点赋值（3）利用四个点坐标确定交点坐标的GetPoint()获得点集Point[][]；（4）利用多边形面积求解方法GetArea()获得面积集合Area[][];（5）取面积集合的最大面积。
3、注意事项：计算几何模板的套用，如求解面积的方法模板需要控制参数点的顺时针或者逆时针顺序；求交点中，用到直线公式注意其K为0和无穷大的情况；给边上点赋值时候注意i，j的变换。
4、实现方法:
#include
<
iostream
>
using
namespace
std;
struct
Node{
double
x,y;
};
int
n;
double
Arr[
5
][
33
],Area[
33
][
33
];
Node Point[
33
][
33
];
//
已知四点坐标求交点
Node GetPoint(Node p1,Node p2,Node p3,Node p4)
{
Node P;
double
A1
=
p2.y
-
p1.y;
double
B1
=
p1.x
-
p2.x;
double
C1
=
p1.y
*
(p2.x
-
p1.x)
-
p1.x
*
(p2.y
-
p1.y);
double
A2
=
p4.y
-
p3.y;
double
B2
=
p3.x
-
p4.x;
double
C2
=
p3.y
*
(p4.x
-
p3.x)
-
p3.x
*
(p4.y
-
p3.y);
P.x
=
(C2
*
B1
-
C1
*
B2)
/
(A1
*
B2
-
A2
*
B1);
P.y
=
(C2
*
A1
-
C1
*
A2)
/
(B1
*
A2
-
B2
*
A1);
return
P;
}
//
获得Point矩阵
void
GetPoints()
{
int
i,j;
//
给四个角的点赋值
Point[
1
][
1
].x
=
0.0
;Point[
1
][
1
].y
=
0.0
;
Point[
1
][n
+
2
].x
=
0.0
;Point[
1
][n
+
2
].y
=
1.0
;
Point[n
+
2
][
1
].x
=
1.0
;Point[n
+
2
][
1
].y
=
0.0
;
Point[n
+
2
][n
+
2
].x
=
1.0
;Point[n
+
2
][n
+
2
].y
=
1.0
;
//
给四条边上的点赋值
for
(i
=
2
;i
<
n
+
2
;i
++
)
{
Point[
1
][i].x
=
0.0
;Point[
1
][i].y
=
Arr[
3
][i
-
1
];
Point[n
+
2
][i].x
=
1.0
;Point[n
+
2
][i].y
=
Arr[
4
][i
-
1
];
Point[i][
1
].x
=
Arr[
1
][i
-
1
];Point[i][
1
].y
=
0.0
;
Point[i][n
+
2
].x
=
Arr[
2
][i
-
1
];Point[i][n
+
2
].y
=
1.0
;
}
for
(i
=
2
;i
<
n
+
2
;i
++
)
for
(j
=
2
;j
<
n
+
2
;j
++
)
Point[i][j]
=
GetPoint(Point[i][
1
],Point[i][n
+
2
],Point[
1
][j],Point[n
+
2
][j]);
}
//
已知4点坐标求四边形面积
double
GetArea(Node p1,Node p2,Node p3,Node p4,
int
N)
{
int
i,j;
double
area
=
0
;
Node polygon[
4
];
polygon[
0
]
=
p1;
polygon[
1
]
=
p2;
polygon[
2
]
=
p3;
polygon[
3
]
=
p4;
for
(i
=
0
;i
<
N;i
++
)
{
j
=
(i
+
1
)
%
N;
area
+=
polygon[i].x
*
polygon[j].y;
area
-=
polygon[i].y
*
polygon[j].x;
}
area
/=
2
;
return
(area
<
0
?
-
area : area);
}
//
获得面积矩阵
void
GetAreas()
{
int
i,j;
for
(i
=
1
;i
<=
n
+
1
;i
++
)
for
(j
=
1
;j
<=
n
+
1
;j
++
)
Area[i][j]
=
GetArea(Point[i][j],Point[i
+
1
][j],Point[i
+
1
][j
+
1
],Point[i][j
+
1
],
4
);
}
double
MaxArea()
{
int
i,j;
double
ans
=
0.0
;
for
(i
=
1
;i
<=
n
+
1
;i
++
)
{
for
(j
=
1
;j
<=
n
+
1
;j
++
)
{
if
(Area[i][j]
>=
ans)
ans
=
Area[i][j];
}
}
return
ans;
}
int
main()
{
while
(cin
>>
n
&&
n)
{
int
i,j;
for
(i
=
1
;i
<=
4
;i
++
)
for
(j
=
1
;j
<=
n;j
++
)
cin
>>
Arr[i][j];
GetPoints();
GetAreas();
printf(
"
%.6f\n
"
,MaxArea());
}
return
0
;
}