2056--求两矩形的交叉面积

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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题目意思就是输入两个矩形的一对对角的顶点坐标，根据这四个坐标求两个矩形的交叉面积。
首先要考虑到这两个矩形是否交叉，判断之后有的话才能进行进一步的计算。而输入时先输入第一个的数据，再输第二个的，需要自己动手考虑到它们之间所有可能的相对位置，很容易漏掉，所以最好亲手画一下图，然后总结计算方式。

#include
double max(double x,double y)
{
    return (x>y)?x:y;
}
double min(double x,double y)
{
    return (xarea,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;
}