2056 Rectangles

Rectangles

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13982    Accepted Submission(s): 4523

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

 

//这里的代码并非原创，我原本是用多个if语句进行判断，后来发现了这种更为简便的算法，觉得简单很多，便引用的过来。
//思路：①四个if语句的作用是将1、3点的坐标分别换为两矩形的左下角顶点的坐标；
  //    ②接下来将1、2点的坐标分别换为两矩形距另一矩形最近的点的坐标；
    //  ③判断两矩形是否交叉，交叉则计算并输出交叉面积，否则输出0。

#include<stdio.h>
double max(double a,double b)
{
    if(a>b)
        return a;
    else
        return b;
}
double min(double a,double b)
{
    if(a<b)
        return a;
    else
        return b;
}
int main()
{
    double x1,y1,x2,y2,x3,y3,x4,y4,t;
    while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
    {
        if (x1>x2) t=x1,x1=x2,x2=t;
        if (y1>y2) t=y1,y1=y2,y2=t;
        if (x3>x4) t=x3,x3=x4,x4=t;
        if (y3>y4) t=y3,y3=y4,y4=t;
            x1=max(x1,x3);
            y1=max(y1,y3);
            x2=min(x2,x4);
            y2=min(y2,y4);
        printf("%.2lf\n",x1>x2||y1>y2?0:(x2-x1)*(y2-y1));
    }
}