HDU_2056——相交矩形的面积

Problem Description

Given two rectangles and the coordinates of two points on the diagonals of each rectangle,you have to calculate the area of the intersected part of two rectangles. its sides are parallel to OX and OY .

 

 

Input

Input The first line of input is 8 positive numbers which indicate the coordinates of four points that must be on each diagonal.The 8 numbers are x1,y1,x2,y2,x3,y3,x4,y4.That means the two points on the first rectangle are(x1,y1),(x2,y2);the other two points on the second rectangle are (x3,y3),(x4,y4).

 

 

Output

Output For each case output the area of their intersected part in a single line.accurate up to 2 decimal places.

 

 

Sample Input

1.00 1.00 3.00 3.00 2.00 2.00 4.00 4.00 5.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50

 

 

Sample Output

1.00 56.25

#include <cstdio>
double max(double a,double b)
{
    return (a>b)?a:b;    
}
double min(double a,double b)
{
    return (a<b)?a:b;    
}
void fun(double &a,double &b)
{
    double t;
    t=a;a=b;b=t;    
}
int main()
{
   double x1,y1,x2,y2,x3,y3,x4,y4;
   while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4))
      {
            //规范对角线坐标，对角线是从左下到右上， 
            if(x1>x2)    fun(x1,x2);
            if(y1>y2)    fun(y1,y2);
            if(x3>x4)    fun(x3,x4);
            if(y3>y4)    fun(y3,y4);
            
            //如果条件不能构成相交矩阵 
            if(x1>=x4 || x2<=x3 || y1>=y4 || y2<=y3)
                {
                    printf("%.2lf\n",0);
                }
            else
                {
                    //构造相交矩阵的对角线坐标(x1,y1),(x2,y2)
                    //相交矩形的对角线坐标能够确定 
                    x1=max(x1,x3);
                    y1=max(y1,y3);
                    x2=min(x2,x4);
                    y2=min(y2,y4);
                    printf("%.2lf\n",(x2-x1)*(y2-y1));
                }
        }
   return 0;
}