HDUOJ 2056 Rectangles (几何计算问题)

Rectangles

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7889    Accepted Submission(s): 2538

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 <algorithm>
#include <iostream>
using namespace std;
int main()
{
	double x[4],y[4],area;
	int i,j;
	while(cin>>x[0]>>y[0]>>x[1]>>y[1]>>x[2]>>y[2]>>x[3]>>y[3])
	{
		//判断两个矩形是否没有公共部分
		for(i=0;i<4;i+=2)
		{
			if(x[i]>x[i+1]) swap(x[i],x[i+1]);
			if(y[i]>y[i+1]) swap(y[i],y[i+1]);
		}
		if(x[2]>=x[1]||x[0]>=x[3]||y[2]>=y[1]||y[0]>=y[3]) area=0;
		//有公共部分时计算面积
		else
		{
			/*排序，排序后x[2]-x[1]即是公共部分的长，
			y[2]-y[1]即是公共部分的高，注意公共部分是矩形*/
			for(i=0;i<4;i++)
			{
				for(j=0;j<3-i;j++)
				{
					if(x[j]>x[j+1])
						swap(x[j],x[j+1]);
					if(y[j]>y[j+1])
						swap(y[j],y[j+1]);
				}
			}

			area=(x[2]-x[1])*(y[2]-y[1]);
		}
		printf("%.2lf\n",area);
	}
	return 0;
}