Fishnet - UVa 1301 几何

A fisherman named Etadokah awoke in a very small island. He could see calm, beautiful and blue sea around the island. The previous night he had encountered a terrible storm and had reached this uninhabited island. Some wrecks of his ship were spread around him. He found a square wood-frame and a long thread among the wrecks. He had to survive in this island until someone came and saved him.

In order to catch fish, he began to make a kind of fishnet by cutting the long thread into short threads and fixing them at pegs on the square wood-frame (Figure 1). He wanted to know the sizes of the meshes of the fishnet to see whether he could catch small fish as well as large ones.

The wood-frame is perfectly square with four thin edges one meter long.. a bottom edge, a top edge, a left edge, and a right edge. There are n pegs on each edge, and thus there are 4n pegs in total. The positions ofpegs are represented by their (x, y) -coordinates. Those of an example case with n = 2 are depicted in Figures 2 and 3. The position of the i th peg on the bottom edge is represented by (a_i, 0) . That on the top edge, on the left edge and on the right edge are represented by (b_i, 1) , (0, c_i) , and (1, d_i) , respectively. The long thread is cut into 2n threads with appropriate lengths. The threads are strained between (a_i, 0) and(b_i, 1) , and between (0, c_i) and (1, d_i) (i = 1,..., n) .

You should write a program that reports the size of the largest mesh among the (n + 1)² meshes of the fishnet made by fixing the threads at the pegs. You may assume that the thread he found is long enough to make the fishnet and that the wood-frame is thin enough for neglecting its thickness.

Input 

The input consists of multiple subproblems followed by a line containing a zero that indicates the end of input. Each subproblem is given in the following format.

n

a₁a₂ ^... a_n

b₁b₂ ^... b_n

c₁c₂ ^... c_n

d₁d₂ ^... d_n

An integer n followed by a newline is the number of pegs on each edge. a₁,..., a_n , b₁,..., b_n , c₁,..., c_n ,d₁,..., d_n are decimal fractions, and they are separated by a space character except that a_n, b_n, c_n and d_n are followed by a new line. Each a_i (i = 1,..., n) indicates the x -coordinate of the i th peg on the bottom edge. Each b_i (i = 1,..., n) indicates the x -coordinate of the i th peg on the top edge. Each c_i (i = 1,..., n)indicates the y -coordinate of the i th peg on the left edge. Each d_i (i = 1,..., n) indicates the y -coordinate of the i th peg on the right edge. The decimal fractions are represented by 7 digits after the decimal point. In addition you may assume that 0 < n30 , 0 < a₁ < a₂ < ^... < a_n < 1 , 0 < b₁ < b₂ < ^... < b_n < 1 , 0 < c₁ < c₂ < ^... < c_n < 1 and 0 < d₁ < d₂ < ^... < d_n < 1 .

Output 

For each subproblem, the size of the largest mesh should be printed followed by a new line. Each value should be represented by 6 digits after the decimal point, and it may not have an error greater than 0.000001.

Sample Input 

20.2000000 0.60000000.3000000 0.80000000.3000000 0.50000000.5000000 0.600000020.3333330 0.66666700.3333330 0.66666700.3333330 0.66666700.3333330 0.666667040.2000000 0.4000000 0.6000000 0.80000000.1000000 0.5000000 0.6000000 0.90000000.2000000 0.4000000 0.6000000 0.80000000.1000000 0.5000000 0.6000000 0.900000020.5138701 0.94762830.1717362 0.17574120.3086521 0.70223130.2264312 0.534534310.40000000.60000000.30000000.50000000

Sample Output 

0.2156570.1111120.0789230.2792230.348958

题意：给你一个（n+1）*（n+1）的网，让你求出最大的四边形的面积。

思路：用几何的两条直线求交点，算出四个端点，然后转化成两个三角形求面积。

AC代码如下：

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y){}};typedef Point Vector;double eps=1e-10;Vector operator+(Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}Vector operator-(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}Vector operator*(Vector A,double p){return Vector(A.x*p,A.y*p);}double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}double area(Point a,Point b,Point c){return abs(Cross(a,b)+Cross(b,c)+Cross(c,a))/2;}Point GetLineInterSection(Point P,Vector v,Point Q,Vector w){    Vector u=P-Q;    double t=Cross(w,u)/Cross(v,w);    return P+v*t;}double a[40],b[40],c[40],d[40];int n;int main(){    int i,j,k;    double ans,ret;    Point A,B,C,D,_A1,_B1,_C1,_D1,_A2,_B2,_C2,_D2;    _A1.y=_A2.y=0;    _B1.y=_B2.y=1;    _C1.x=_C2.x=0;    _D1.x=_D2.x=1;    while(~scanf("%d",&n) && n>0)    {        for(i=1;i<=n;i++)           scanf("%lf",&a[i]);        for(i=1;i<=n;i++)           scanf("%lf",&b[i]);        for(i=1;i<=n;i++)           scanf("%lf",&c[i]);        for(i=1;i<=n;i++)           scanf("%lf",&d[i]);        n++;        a[0]=b[0]=c[0]=d[0]=0;        a[n]=b[n]=c[n]=d[n]=1;        ans=0;        for(i=1;i<=n;i++)           for(j=1;j<=n;j++)           {               _A1.x=a[j-1];_A2.x=a[j];               _B1.x=b[j-1];_B2.x=b[j];               _C1.y=c[i-1];_C2.y=c[i];               _D1.y=d[i-1];_D2.y=d[i];               A=GetLineInterSection(_A1,_B1-_A1,_C1,_D1-_C1);               B=GetLineInterSection(_A2,_B2-_A2,_C1,_D1-_C1);               C=GetLineInterSection(_A2,_B2-_A2,_C2,_D2-_C2);               D=GetLineInterSection(_A1,_B1-_A1,_C2,_D2-_C2);               ret=area(A,B,C)+area(A,D,C);               ans=max(ans,ret);           }        printf("%.6f\n",ans);    }}