Rectangles
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19110 Accepted Submission(s): 6187
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
解析：调整坐标，找出重复部分的长和宽，从而求得面积。
1 #include <cstdio> 2 #include <algorithm> 3 using namespace std; 4 5 double x1,y1,x2,y2,x3,y3,x4,y4; 6 double tmp; 7 8 int main() 9 { 10 while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)){ 11 if(x1>x2){ //使(x1,y1)为左下角点,(x2,y2)为右上角点 12 tmp = x1; 13 x1 = x2; 14 x2 = tmp; 15 } 16 if(y1>y2){ 17 tmp = y1; 18 y1 = y2; 19 y2 = tmp; 20 } 21 if(x3>x4){ //使(x3,y3)为左下角点,(x4,y4)为右上角点 22 tmp = x3; 23 x3 = x4; 24 x4 = tmp; 25 } 26 if(y3>y4){ 27 tmp = y3; 28 y3 = y4; 29 y4 = tmp; 30 } 31 double length = min(x2,x4)-max(x1,x3); 32 double width = min(y2,y4)-max(y1,y3); 33 printf("%.2f ",length<0 || width<0 ? 0:length*width); 34 } 35 return 0; 36 }