2056 Rectangles

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

参考链接：HDU  2056  (矩形重叠)Rectangles

最终AC代码：

#include <bits/stdc++.h>
using namespace std;
//难点在于判断是否有重叠！方法是：水平方向：取两个矩形左边大者记为a；
//取两个矩形右边小者记为b；若a > b则说明没有重叠部分。 垂直方向同理！ 
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)!=EOF)
   {
       //转换为正对角线坐标 因为测试用例可能输入副对角线的点 
       if(x1>x2) swap(x1, x2);
       if(y1>y2) swap(y1, y2);
       if(x3>x4) swap(x3, x4);
       if(y3>y4) swap(y3, y4);
       //找到重叠的部分 
       x1 = max(x1, x3); 
       y1 = max(y1, y3);
       x2 = min(x2, x4);
       y2 = min(y2, y4);
       printf("%.2lf
", (x1>x2||y1>y2)?0:(x2-x1)*(y2-y1));
    }
}