poj 1408 Fishnet（计算几何 叉积求面积 求两直线交点 暴力）

Fishnet
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 2103
Accepted: 1325
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(遭遇) a terrible storm and had reached this uninhabited(无人居住的) island. Some wrecks(破坏) of his ship were spread around him. He found a square wood-frame(木制结构) and a long thread among the wrecks. He had to survive(幸存) in this island until someone came and saved him.


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


The wood frame(框架) 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(描述) 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(分别地). The long thread is cut into 2n threads with appropriate(适当的) lengths. The threads are strained(拉紧) 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(承担) that the thread he found is long enough to make the fishnet and the wood-frame is thin enough for neglecting(疏忽) its thickness(厚度).


Input
The input(投入) consists of multiple sub-problems followed by a line containing a zero that indicates(表明) 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(承担) 0 < n <= 30, 0 < ai,bi,ci,di < 1
Output
For each sub-problem, the size of the largest mesh(网眼) should be printed followed by a new line. Each value should be represented by 6 digits(数字) after the decimal(小数) 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

计算两条直线交点

//计算两直线交点,注意事先判断直线是否平行!
//线段交点请另外判线段相交(同时还是要判断是否平行!)12
point intersection(point u1,point u2,point v1,point v2)
{
point ret=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ret.x+=(u2.x-u1.x)*t;
ret.y+=(u2.y-u1.y)*t;
return ret;
}

#include<stdio.h>
#include<math.h>


struct node
{
    double x,y;
}point[35][35];
int n;
void init()//初始化
{
    point[0][0].x=0;
    point[0][0].y=0;


    point[0][n+1].x=1.0;
    point[0][n+1].y=0;


    point[n+1][0].x=0;
    point[n+1][0].y=1.0;


    point[n+1][n+1].x=1.0;
    point[n+1][n+1].y=1.0;
}
struct node get_point(struct node a,struct node b,struct node c,struct node d)//求两条直线的交点
{
    struct node 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;
};


double get_area(struct node a,struct node b,struct node c)//叉积求面积
{
    return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);
}


int main()
{


    int i,j,k;
    while(~scanf("%d",&n),n)
    {
        init();
        for(i=1;i<=n;i++)
        {
            scanf("%lf",&point[0][i].x);
            point[0][i].y=0;
        }


         for(i=1;i<=n;i++)
        {
            scanf("%lf",&point[n+1][i].x);
            point[n+1][i].y=1.0;
        }


         for(i=1;i<=n;i++)
        {
            scanf("%lf",&point[i][0].y);
            point[i][0].x=0;
        }


         for(i=1;i<=n;i++)
        {
            scanf("%lf",&point[i][n+1].y);
            point[i][n+1].x=1.0;
        }


        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
        {
               point[i][j]=get_point(point[0][j],point[n+1][j],point[i][0],point[i][n+1]);
        }
        double area  = 0;
        double tmp;
        for(i=1;i<=n+1;i++)
        {
            for(j=1;j<=n+1;j++)
            {
                tmp=fabs(get_area(point[i-1][j-1],point[i][j],point[i-1][j]));//四边形面积分为两个三角形面积（注意正负号）
                tmp+=fabs(get_area(point[i-1][j-1],point[i][j],point[i][j-1]));
                tmp/=2;
                if(tmp>area)
                    area=tmp;
            }
        }
         printf("%.6f\n",area);
    }






return 0;


}