Rectangles

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

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即可

这道题不难 可是我却 用了整个中午的时间搞他

原先代码不知是哪的毛病 測试数据也对 但就是

wa

后来没有办法又一次换了一种算法 这才算对

代码例如以下：

#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)//推断不重叠的时候 为零由于题目有要求是精确到两位小数

s=0.00;

printf("%.2lf\n",s);

}

return 0;

}