5D

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

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

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

// 没考虑无相交区域

 1 #include<stdio.h>
 2 int main()
 3 {
 4     double x1,y1, x2,y2, x3,y3, x4,y4, a,b, t;
 5     while(scanf("%lf %lf %lf %lf %lf %lf %lf %lf", &x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
 6     {
 7         if(x1>x2)
 8         { t=x2; x2=x1; x1=t; }
 9         if(y1>y2)
10         { t=y2; y2=y1; y1=t; }
11         if(x3>x4)
12         { t=x4; x4=x3; x3=t; }
13         if(y3>y4)
14         { t=y4; y4=y3; y3=t; }
15         a=x1-x4>x3-x2?x1-x4:x3-x2;
16         b=y1-y4>y3-y2?y1-y4:y3-y2;
17         printf("%.2f
", a*b);
18     }
19     return 0;
20 }

WA

// 还是错得离谱 感觉没有智商T^T

 1 #include<stdio.h>
 2 int main()
 3 {
 4     double x1,y1, x2,y2, x3,y3, x4,y4, a,b, t;
 5     while(scanf("%lf %lf %lf %lf %lf %lf %lf %lf", &x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
 6     {
 7         if(x1>x2)
 8         { t=x2; x2=x1; x1=t; }
 9         if(y1>y2)
10         { t=y2; y2=y1; y1=t; }
11         if(x3>x4)
12         { t=x4; x4=x3; x3=t; }
13         if(y3>y4)
14         { t=y4; y4=y3; y3=t; }
15         a=x1-x4>x3-x2?x1-x4:x3-x2;
16         b=y1-y4>y3-y2?y1-y4:y3-y2;
17         if(a>=0||b>=0) printf("0.00
");
18         else printf("%.2f
", a*b);
19     }
20     return 0;
21 }

WA*2

//

 1 #include<stdio.h>
 2 int main()
 3 {
 4     double x1,y1, x2,y2, x3,y3, x4,y4, a,b, t;
 5     while(scanf("%lf %lf %lf %lf %lf %lf %lf %lf", &x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
 6     {
 7         if(x1>x2)
 8         { t=x2; x2=x1; x1=t; }
 9         if(y1>y2)
10         { t=y2; y2=y1; y1=t; }
11         if(x3>x4)
12         { t=x4; x4=x3; x3=t; }
13         if(y3>y4)
14         { t=y4; y4=y3; y3=t; }
15         a=(x2<x4?x2:x4)-(x1>x3?x1:x3);
16         b=(y2<y4?y2:y4)-(y1>y3?y1:y3);
17         if(a<0||b<0) printf("0.00
");
18         else printf("%.2f
", a*b);
19     }
20     return 0;
21 }

AC

原文地址：https://www.cnblogs.com/goldenretriever/p/10355783.html