HDU 2056 Rectangles（計算相交面積）

Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20183Accepted Submission(s): 6537

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
Author seeyou
Source 校慶杯Warm Up 題解：求出兩個長方形相交的面積，如果重合的話就為0。
AC代碼：

#include
#include
#include
#include
#include
#include
#include
#include
#include
typedef long long LL;
using namespace std;
int main()
{
	
	double x1,y1,x2,y2,x3,y3,x4,y4;
	while(cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4)
	{
	//保證左邊是右下角，右邊的數是左上角 (畫個圖吧...)
		if(x1>x2)swap(x1,x2); 
		if(y1>y2)swap(y1,y2);
		
		if(x3>x4)swap(x3,x4);
		if(y3>y4)swap(y3,y4);
    //畫個圖就知道是什麼意思了 
		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));//注意重合的情況 
	}
    return 0;
}