矩形 2056

Rectangles

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19475    Accepted Submission(s): 6316

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

 

注： 刚开始感觉这道题不难，但却不好做，需要考虑分析的情况太多了。思考了好久才找到下面本人认为比较巧妙的方法。可是在提交代码却wrong answer想了好久才发现自己忽略了下图中的第四种情况。

My   solution:

   
   
#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
int i,j,k,n,m;
double x[ 5],y[ 5],a[ 5],b[ 5],s;
while( scanf( "%lf%lf%lf%lf%lf%lf%lf%lf",&x[ 1],&y[ 1],&x[ 2],&y[ 2],&x[ 3],&y[ 3],&x[ 4],&y[ 4])== 8)
{
s= 0;
for(i= 1;i<= 4;i++)
{
a[i]=x[i];
b[i]=y[i];
} //a[]、b[]数组记录原来数据
sort(x+ 1,x+ 5);
sort(y+ 1,y+ 5);
if(x[ 3]>a[ 2]&&x[ 3]>a[ 1]||x[ 3]>a[ 3]&&x[ 3]>a[ 4]||y[ 3]>b[ 1]&&y[ 3]>b[ 2]||y[ 3]>b[ 3]&&y[ 3]>b[ 4])
; //判断两个矩形没有相交
else
s=(x[ 3]-x[ 2])*(y[ 3]-y[ 2]);
printf( "%.2lf\n",s);
}

return 0;
}