HDU 2056 Rectangles

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\n",length<0 || width<0 ? 0:length*width);
34     }
35     return 0;
36 }

 

 

转载于:https://www.cnblogs.com/inmoonlight/p/5165754.html