Problem 1009

Problem 1009
Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 81  Solved: 34
[Submit][Status][Web Board]
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

标程：

#include <cstdio> 
#include <cstring> 
#include <cstdlib> 
using namespace std; 
  
struct point{ 
   double x,y; 
   int node; 
}temp[4]; 
  
int  cmp(const void*a,const void *b) 
{ 
    struct point *aa=(point*)a; 
    struct point *bb=(point*)b; 
    return(((aa->x)>(bb->x))?1:-1); 
} 
int recmp(const void *a,const void *b) 
{ 
    struct point*aa=(point*)a; 
    struct point *bb=(point *)b; 
    return (((aa->y)>(bb->y))?1:-1); 
} 
double area() 
{ 
    qsort(temp,4,sizeof(temp[0]),cmp); 
    double temp1; 
    if(temp[0].node==temp[1].node) temp1=0; 
    else temp1=temp[2].x-temp[1].x; 
    qsort(temp,4,sizeof(temp[0]),recmp); 
    double temp2; 
    if(temp[0].node==temp[1].node) temp2=0; 
    else temp2=temp[2].y-temp[1].y; 
    return temp1*temp2; 
} 
int main() 
{ 
    //freopen("in.txt","r",stdin); 
    //freopen("out.txt","w",stdout); 
  while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&temp[0].x,&temp[0].y, &temp[1].x,&temp[1].y,&temp[2].x,&temp[2].y,&temp[3].x,&temp[3].y)!=EOF) 
  { 
      temp[1].node=temp[0].node=1; 
      temp[3].node=temp[2].node=2; 
      printf("%.2lf\n",area()); 
  } 
  return 0; 
}