POJ 1408 Fishnet (判断围成四边形最大面积，直线相交问题)

Fishnet

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1935		Accepted: 1242

Description

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. 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 on 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 of pegs are represented by their (x,y)-coordinates. Those of an example case with n=2 are depicted in figures below. The position of the ith peg on the bottom edge is represented by (ai,0). That on the top edge, on the left edge and on the right edge are represented by (bi,1), (0,ci) and (1,di), respectively. The long thread is cut into 2n threads with appropriate lengths. The threads are strained between (ai,0) and (bi,1),and between (0,ci) and (1,di) (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 the wood-frame is thin enough for neglecting its thickness.

Input

The input consists of multiple sub-problems followed by a line containing a zero that indicates the end of input. Each sub-problem is given in the following format.
n
a1 a2 ... an
b1 b2 ... bn
c1 c2 ... cn
d1 d2 ... dn
you may assume 0 < n <= 30, 0 < ai,bi,ci,di < 1

Output

For each sub-problem, 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

2
0.2000000 0.6000000
0.3000000 0.8000000
0.1000000 0.5000000
0.5000000 0.6000000
2
0.3333330 0.6666670
0.3333330 0.6666670
0.3333330 0.6666670
0.3333330 0.6666670
4
0.2000000 0.4000000 0.6000000 0.8000000
0.1000000 0.5000000 0.6000000 0.9000000
0.2000000 0.4000000 0.6000000 0.8000000
0.1000000 0.5000000 0.6000000 0.9000000
2
0.5138701 0.9476283
0.1717362 0.1757412
0.3086521 0.7022313
0.2264312 0.5345343
1
0.4000000
0.6000000
0.3000000
0.5000000
0

Sample Output

0.215657
0.111112
0.078923
0.279223
0.348958



题意：有一个1*1的正方形在四条边上各取n个点，然后求在正方形内围成的最大四边形的面积

思路：用一个矩阵来保存所有的点，四边上每个点是输入的，内部的每个点通过线段交点的计算可以计算出来。
然后枚举任意i-1,i,j-1,j四个点计算四边形的面积，求最大值。在计算四边形面积的时候四边形可以转换成两
个三角形来计算，这两个三角形的面积是通过向量的叉积来计算的。两个向量的叉积可以算出以这两个向量为
邻边的四边形的面积，注意除以2.

ac代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define fab(a) (a)>0?(a):(-a)
#define LL long long
#define MAXN 2100
#define mem(x) memset(x,0,sizeof(x))
#define INF 0xfffffff 
#define PI acos(-1.0)
using namespace std;
struct s
{
	double x,y;
}list[MAXN][MAXN];
double dis(s aa,s bb)
{
	return sqrt((aa.x-bb.x)*(aa.x-bb.x)+(aa.y-bb.y)*(aa.y-bb.y));
}
double area(s a,s b,s c)//叉积求面积
{  
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);  
}
s fun(s a,s b,s c,s d)//求交点
{
    s temp=a;
    double t=((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));
    temp.x+=(b.x-a.x)*t;
    temp.y+=(b.y-a.y)*t;
    return temp;
}
void intn(int n)
{
	double a;
	int i,j;
	list[0][0].x=0,list[0][0].y=0;
	list[0][n+1].x=1.0,list[0][n+1].y=0.0;
	list[n+1][n+1].x=1.0,list[n+1][n+1].y=1.0;
	list[n+1][0].x=0.0,list[n+1][0].y=1.0;
	for(i=1;i<=n;i++){
		scanf("%lf",&a);
		list[0][i].x=a,list[0][i].y=0.0;
	}
	for(i=1;i<=n;i++){
		scanf("%lf",&a);
		list[n+1][i].x=a,list[n+1][i].y=1.0;
	}
	for(i=1;i<=n;i++){
		scanf("%lf",&a);
		list[i][0].x=0.0,list[i][0].y=a;
	}
	for(i=1;i<=n;i++){
		scanf("%lf",&a);
		list[i][n+1].x=1.0,list[i][n+1].y=a;
	}
}
int main()
{
	int n,i,j;
	while(scanf("%d",&n)!=EOF,n)
	{
		intn(n);
		for(j=1;j<=n;j++)
        {
            for(i=1;i<=n;i++)
            {
             list[i][j]=fun(list[0][j],list[n+1][j],list[i][0],list[i][n+1]);
            }
        }
        double M=0.0,ans;
        for(i=1; i<=n+1; i++)
        {
            for(j=1; j<=n+1; j++)
            {
                ans=fab(area(list[i-1][j-1],list[i][j],list[i][j-1]));
                ans+=fab(area(list[i-1][j-1],list[i][j],list[i-1][j]));
                ans/=2;
                M=max(M,ans);
            }
        }
        printf("%.6lf\n",M);
	}
	return 0;
}