Fishnet(几何)

http://poj.org/problem?id=1408

题意：给出 a1 a2 ... an
               b1 b2 ... bn 
               c1 c2 ... cn 
               d1 d2 ... dn 这些点，求这些对应点连线形成的小四边形的最大面积。

思路：将所有的交点求出，同已知点一起存入二维矩阵中，枚举每个小四边形，求出其面积，找出最大的即可。

 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
 const double eps=1e-;//设置精度
 #define max1(a1,b1) (double)a1-(double)b1>eps?(double)a1:(double)b1
 struct point//点
 {
     double x;
     double y;
 } map[][];//用二维数组的结构体存储所有的点
 struct line//线
 {
     double a,b,c;
 };
 line getline(point p1,point p2)//由两点求直线ax+by+c=0
 {
     line tmp;
     tmp.a = p1.y-p2.y;
     tmp.b = p2.x-p1.x;
     tmp.c = p1.x*p2.y-p2.x*p1.y;
     return tmp;
 }
 point getIntersect(line L1,line L2)//求两直线交点
 {
     point tmp;
     tmp.x = (L1.b*L2.c-L2.b*L1.c)/(L1.a*L2.b-L2.a*L1.b);
     tmp.y = (L1.c*L2.a-L2.c*L1.a)/(L1.a*L2.b-L2.a*L1.b);
     return tmp;
 }
 double area_polygon(int n,point* p)//求多边形面积
 {
     double s1=,s2=;
     int i;
     for (i=; i<n; i++)
     {
         s1+=p[(i+)%n].y*p[i].x;
         s2+=p[(i+)%n].y*p[(i+)%n].x;
     }
     return fabs(s1-s2)/;
 }
 int main()
 {
     int n;
     while(~scanf("%d",&n)&&n)
     {
         double Max = eps;
         double a,b,c,d;
         map[][].x = ;//初始化已知的四个顶点的坐标
         map[][].y = ;
         map[][n+].x = ;
         map[][n+].y = ;
         map[n+][].x = ;
         map[n+][].y = ;
         map[n+][n+].x = ;
         map[n+][n+].y = ;
         for (int j = ; j <= n; j++)//输入a1 a2 ... an，并存储其坐标
         {
             scanf("%lf",&a);
             map[][j].x = a;
             map[][j].y = ;
         }
         for (int j = ; j <= n; j++)//输入b1 b2 ... bn，并存储其坐标
         {
             scanf("%lf",&b);
             map[n+][j].x = b;
             map[n+][j].y = ;
         }
         for (int i = ; i <= n; i++)//输入c1 c2 ... cn，并存储其坐标
         {
             scanf("%lf",&c);
             map[i][].x = ;
             map[i][].y = c;
         }
         for (int i =; i <= n; i++)//输入d1 d2 ... dn，并存储其坐标
         {
             scanf("%lf",&d);
             map[i][n+].x = ;
             map[i][n+].y = d;
         }
         for (int i = ; i <= n; i++)
             for (int j = ; j <= n; j++)
             {
                 line L1 = getline(map[i][],map[i][n+]);//枚举所有可相交的直线
                 line L2 = getline(map[][j],map[n+][j]);
                 point tmp = getIntersect(L1,L2);//求这两条线的交点
                 map[i][j].x = tmp.x;//将求得的交点坐标存入相应的数组中
                 map[i][j].y = tmp.y;

             }
         point p[];//存储四边形的顶点
         for (int i = ; i <= n+; i++)
         {

             for (int j = ; j <= n+; j++)//枚举所有四边形右上角的顶点
             {
                 p[] = map[i][j];//以map[i][j]为右上角的顶点组成的四边形的各点
                 p[] = map[i][j-];
                 p[] = map[i-][j-];
                 p[] = map[i-][j];
                 double s = area_polygon(,p);//求四边形的面积
                 Max = max1(s,Max);//求最大面积
             }

         }
         printf("%.6f\n",Max);
     }
     return ;
 }