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.
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題目意思就是輸入兩個矩形的一對對角的頂點座標，根據這四個座標求兩個矩形的交叉面積。
首先要考慮到這兩個矩形是否交叉，判斷之後有的話才能進行進一步的計算。而輸入時先輸入第一個的數據，再輸第二個的，需要自己動手考慮到它們之間所有可能的相對位置，很容易漏掉，所以最好親手畫一下圖，然後總結計算方式。
#include<stdio.h>
double max(double x,double y)
{
return (x>y)?x:y;
}
double min(double x,double y)
{
return (x<y)?x:y;
}
int main()
{
double x1,x2,x3,x4,y1,y2,y3,y4,area,a,b;
int flag=0;
while (scanf ("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
{
if (min(x1,x2)>=max(x3,x4)||min(y1,y2)>=max(y3,y4)
||max(x1,x2)<=min(x3,x4)||max(y1,y2)<=min(y3,y4))
printf ("0.00\n");
else
{
a=min(max(x3,x4),max(x1,x2))-max(min(x3,x4),min(x1,x2));
b=min(max(y3,y4),max(y1,y2))-max(min(y3,y4),min(y1,y2));
area=a*b;
if (area<0)
area=(-1)*area;
printf ("%.2lf\n",area);
}
}
return 0;
}