poj 1408 Fishnet(计算几何)

题意：给出一个1*1的正方形，每条边上都有n个点，由对边上的点连成的线段将正方形分成(n+1)*2个不规则四边形，求其中面积最大的面积。

思路：利用叉积求出所有焦点，然后用叉积求四边形面积求的所有的面积找最大的一个～

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#define  N 34
using namespace std;

struct node
{
double x , y ;
}p

;

int n ;
//叉积求交点
struct node point ( struct node p1 , struct node p2 , struct node p3 , struct node p4 )
{
struct node td ;
double A1=p2.y-p1.y;
double B1=p1.x-p2.x;
double C1=p1.y*(p2.x-p1.x)-p1.x*(p2.y-p1.y);
double A2=p4.y-p3.y;
double B2=p3.x-p4.x;
double C2=p3.y*(p4.x-p3.x)-p3.x*(p4.y-p3.y);
td.x=(C2*B1-C1*B2)/(A1*B2-A2*B1);
td.y=(C2*A1-C1*A2)/(B1*A2-B2*A1);
return td ;
}
//计算面积
double area ( struct node p1 , struct node p2 , struct node p3 , struct node p4 )
{
int i , j ;
struct node are[4] ;
are[0] = p1 ;
are[1] = p2 ;
are[2] = p3 ;
are[3] = p4 ;

double sum = 0.0 ;
for ( i = 0 ; i < 4 ; i++ )
{
j = ( i + 1 ) % 4 ;
sum += are[i].x * are[j].y - are[i].y * are[j].x ;
}
sum /= 2 ;
if ( sum < 0 )
sum = -sum ;
return sum ;
}

int main()
{
int i , j , k ;

while ( scanf ( "%d" , &n) , n )
{
//给四个角赋值
p[0][0].x = 0.0 ;p[0][0].y = 0.0 ;
p[0][n+1].x = 0.0 ; p[0][n+1].y = 1.0 ;
p[n+1][n+1].x = 1.0 ; p[n+1][n+1].y = 1.0 ;
p[n+1][0].x = 1.0 ; p[n+1][0].y = 0.0 ;
//输入四条边上的点
for ( i = 1 ; i <= n ; i++ )
{
scanf ( "%lf" , &p[i][0].x) ;
p[i][0].y = 0.0 ;
}
for ( i = 1 ; i <= n ; i++ )
{
scanf ( "%lf" , &p[i][n+1].x) ;
p[i][n+1].y = 1.0 ;
}
for ( i = 1 ; i <= n ; i++ )
{
scanf ( "%lf" , &p[0][i].y) ;
p[0][i].x = 0.0 ;
}
for ( i = 1 ; i <= n ; i++ )
{
scanf ( "%lf" , &p[n+1][i].y) ;
p[n+1][i].x = 1.0 ;
}
//求出其他点
for ( i = 1 ; i <= n ; i++ )
{
for ( j = 1 ; j <= n ; j++ )
p[i][j] = point ( p[i][0] , p[i][n+1] , p[0][j] , p[n+1][j] ) ;
}
/*for ( i = n+1 ; i >= 0 ; i-- )
{
for ( j =0 ; j <= n+1 ; j++ )
printf ( "(%lf %lf)" , p[j][i].x , p[j][i].y );
printf ( "\n" );
}*/
//求出最大面积
double maxx = -1.0 ;
for ( i = 0 ; i < n + 1 ; i++ )
{
for ( j = 0 ; j < n + 1 ; j++ )
{
double sum = area( p[i][j] , p[i+1][j] , p[i+1][j+1] , p[i][j+1] ) ;
if ( sum > maxx )
maxx = sum ;
}
}
printf ( "%.6lf\n" , maxx );
}
return 0 ;
}