HDU2056 rectangles "rectangular area intersection"

RectanglesTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others) Total Submission (s): 15950 Accepted Submission (s): 5104

Problem Descriptiongiven-rectangles and the coordinates of the points on the diagonals of each rectangle,you has to C Alculate the area of the intersected part of the rectangles. Its sides is parallel to OX and OY.
Inputinput the first line of input was 8 positive numbers which indicate the coordinates of four points that must being on EAC H Diagonal. The 8 numbers is x1,y1,x2,y2,x3,y3,x4,y4. That means the points on the first rectangle is (x1,y1), (x2,y2); the other and the points on the second rectangle is (x3,y 3), (X4,Y4).
Outputoutput for each case output, the area of their intersected part, 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 4.005.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50

Sample Output

1.0056.25

find the upper and lower right coordinates of the overlapping rectangles, the code is a bit messy

#include <stdio.h> #include <math.h>struct Node {double x1, y1, x2, y2;} p1, P2;double min (double A, double b) {return a < b? A:b;} Double Max (double A, double b) {return a > b a:b;} void Swap (double& A, double& b) {Double t = A; a = b; b = t;} int main () {freopen ("Stdin.txt", "R", stdin);d ouble x1, y1, x2, Y2;while (scanf ("%lf%lf%lf%lf%lf%lf%lf%lf", &p1.x1, & Amp;p1.y1, &p1.x2, &p1.y2, &p2.x1, &p2.y1, &p2.x2, &p2.y2)! = EOF) {x1 = min (p1.x1, p1.x2); x2 = ma X (p1.x1, p1.x2); y1 = min (p1.y1, p1.y2); y2 = max (p1.y1, p1.y2);p 1.x1 = X1;p1.y1 = y1;p1.x2 = X2;p1.y2 = y2;x1 = Min (p2.x1,  P2.X2); x2 = max (p2.x1, p2.x2); y1 = min (p2.y1, p2.y2); y2 = max (p2.y1, p2.y2);p 2.x1 = X1;p2.y1 = y1;p2.x2 = X2;p2.y2 = y2;x1  = Max (p1.x1, p2.x1); y1 = max (p1.y1, p2.y1); x2 = min (p1.x2, p2.x2); y2 = min (p1.y2, p2.y2);p rintf ("%.2lf\n", x1 >= x2 | | Y1 >= y2? 0.0: (X2-X1) * (y2-y1));} return 0;}

HDU2056 rectangles "rectangular area intersection"