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杭电 2056

2024-07-14 20:35:36   214人阅读

Rectangles

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14340    Accepted Submission(s): 4629


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 4.005.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50
 

Sample Output
1.0056.25
 

Author
seeyou
 

Source
校庆杯Warm Up
 
这道题就是让求两个矩形重叠的面积用长乘宽就行
这道题不难 但是我却 用了整个中午的时间搞他
原先代码不知是哪的毛病 测试数据也对 但就是
wa
后来没有办法重新换了一种算法 这才算对
代码如下：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
int cmp(const void *a,const void *b)
{
 return *(double *)a>*(double *)b?1:-1;
}
int main()
{
 double s,l,h;
 double x[4],y[4];
 double x1,y1,x2,x3,x4,y2,y3,y4;
 while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4))
 {
  x[0]=x1;x[1]=x2;x[2]=x3;x[3]=x4;
  y[0]=y1;y[1]=y2;y[2]=y3;y[3]=y4;
  qsort(x,4,sizeof(x[0]),cmp);
  qsort(y,4,sizeof(y[0]),cmp);
  l=fabs(x2-x1)+fabs(x4-x3)-(x[3]-x[0]);
  h=fabs(y2-y1)+fabs(y4-y3)-(y[3]-y[0]);
  s=l*h;
  if(l<=0||h<=0)//判断不重叠的时候 为零因为题目有要求是精确到两位小数 
  s=0.00;
  printf("%.2lf\n",s);
 }
 return 0;
}

 


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