2013
12-26

# Rectangles

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 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).

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).

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

1.00
56.25

简单数学题，但是细节就太重要了，注意面积为零的情况，最最该注意的是题中给出的对角线上的点是主对角线或副对角线上的，我就是因为一直认为它给出的都是主对角线上的点错了好多次。基本步骤是先将给出的点转换成主对角线上的点，即用主对角线的两个点表示矩形，再确定重叠部分（假设存在）矩形的主对角线的两个点坐(x1,y1)(x2,y2)如果x2-x1<0 或 y2-y1<0，那么假设不成了，重叠部分为零（输出时应输出0.00）。

#include <stdio.h>

void change(double *x1, double *y1, double *x2, double *y2)
{//将题中给出的矩形，用主对角线的顶点表示
double t;
if (*x1 > *x2)
{
t = *x1;
*x1 = *x2;
*x2 = t;
}
if (*y1 > *y2)
{
t = *y1;
*y1 = *y2;
*y2 = t;
}
}

int main()
{
double A[8], x1, y1, x2, y2;
while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",	&A[0], &A[1], &A[2], &A[3],	&A[4], &A[5], &A[6], &A[7]) != EOF)
{
change (&A[0], &A[1], &A[2], &A[3]);
change (&A[4], &A[5], &A[6], &A[7]);
x1 = A[0] > A[4] ? A[0] : A[4];//
y1 = A[1] > A[5] ? A[1] : A[5];//
x2 = A[2] < A[6] ? A[2] : A[6];//
y2 = A[3] < A[7] ? A[3] : A[7];//求出重叠部分（假定重叠）的主对角线顶点
if( (x2 - x1) < 0 || (y2 - y1) < 0)
printf("0.00\n");
else
printf("%.2lf\n", (x2-x1)*(y2-y1));
}
return 0;
}

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