1 /*
2 LA2402：
3 题意：在矩形中给定横线（竖）之间不交叉的n对线，求被分割的小块的最大面积
4 读懂题意就可以发现是思路很简单的题目
5 枚举+几何计算，时间复杂度度不高
6 熟悉了部分函数的运用
7
8 */
9 #include <stdio.h>
10 #include <stdlib.h>
11 #include <string.h>
12 #include <math.h>
13 #include <ctype.h>
14 #include <string>
15 #include <iostream>
16 #include <sstream>
17 #include <vector>
18 #include <queue>
19 #include <stack>
20 #include <map>
21 #include <list>
22 #include <set>
23 #include <algorithm>
24
25 using namespace std;
26
27 struct Point
28 {
29 double x,y;
30 Point(double x=0,double y=0):x(x),y(y){}
31 void print()
32 {
33 cout<<"("<<x<<","<<y<<")"<<endl;
34 }
35 };
36
37 typedef Point Vector;
38
39 Vector operator-(Point A,Point B)//表示A指向B
40 {
41 return Vector(A.x-B.x,A.y-B.y);
42 }
43 Vector operator*(Vector A,double k)
44 {
45 return Vector(A.x*k,A.y*k);
46 }
47 Vector operator+(Point A,Point B)//表示A指向B
48 {
49 return Vector(B.x+A.x,B.y+A.y);
50 }
51 double Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;}
52 double Area(Point A,Point B,Point C)//三角形面积
53 {
54 return fabs(Cross(B-A,C-A))/2;
55 }
56 struct Line
57 {
58 Point p;
59 Vector v;
60 Line(Point p,Vector v):p(p),v(v){}
61 };
62 Line Getline(Point A,Point B)//求直线AB
63 {
64 Vector u=A-B;
65 return Line(A,u);
66 }
67 Point InterSection(Line L1,Line L2)//求直线交点
68 {
69 Vector u=L1.p-L2.p;
70 double t=Cross(L2.v,u)/Cross(L1.v,L2.v);
71 return L1.p+L1.v*t;
72 }
73 double a[50],b[50],c[50],d[50];
74 int n;
75 int main()
76 {
77 while(cin>>n)
78 {
79 if(!n) break;
80 for(int i=1;i<=n;i++) cin>>a[i];
81 for(int i=1;i<=n;i++) cin>>b[i];
82 for(int i=1;i<=n;i++) cin>>c[i];
83 for(int i=1;i<=n;i++) cin>>d[i];
84 a[0]=0;b[0]=0;c[0]=0;d[0]=0;
85 a[n+1]=1;b[n+1]=1;c[n+1]=1;d[n+1]=1;
86 double m=-1;
87 for(int i=0;i<=n;i++)//外层枚举相邻的横线
88 {
89 Line L1=Getline(Point(0,c[i]),Point(1,d[i]));
90 Line L2=Getline(Point(0,c[i+1]),Point(1,d[i+1]));
91 for(int j=0;j<=n;j++)//内层枚举相邻的竖线
92 {
93 Line L3=Getline(Point(a[j],0),Point(b[j],1));//一开始写成i，= =！
94 Line L4=Getline(Point(a[j+1],0),Point(b[j+1],1));
95 Point P1=InterSection(L1,L3);
96 Point P2=InterSection(L1,L4);
97 Point P3=InterSection(L2,L4);
98 Point P4=InterSection(L2,L3);
99 double S=Area(P1,P2,P4)+Area(P2,P3,P4);
100 // P1.print();P2.print();P3.print();P4.print();
101 // cout<<"A1="<<Area(P1,P2,P4)<<endl;
102 // cout<<"A2="<<Area(P2,P3,P4)<<endl;
103 // cout<<"S="<<S<<endl;
104 if (S>m) m=S;
105 }
106 }
107 printf("%.6lf\n",m);
108 }
109 return 0;
110 }
