題目大意：給兩個矩形對角線上的點(主對角線或者副對角線)，求兩個矩形重疊部分的面積。
輸入：每一行都是八個數字，對應4個點。
輸出：輸出重疊的面積，精確到小數點後兩位。
方法：
/*
*
*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
*校慶杯Warm Up
*
*
*Recommend
*linle
*
*/
#include<iostream>
using namespace std;
// 0 1 2 3 4 5 6 7
double a[8];// x1,y1,x2,y2,x3,y3,x4,y4
double point[4];
int i;
int main() {
while (~scanf_s("%lf%lf%lf%lf%lf%lf%lf%lf", &a[0],&a[1], &a[2], &a[3], &a[4], &a[5], &a[6], &a[7])){
if (a[0] > a[2]) swap(a[0], a[2]);
if (a[1] > a[3]) swap(a[1], a[3]);
if (a[4] > a[6]) swap(a[4], a[6]);
if (a[5] > a[7]) swap(a[5], a[7]);
point[0] = a[0] > a[4] ? a[0] : a[4];
point[1] = a[1] > a[5] ? a[1] : a[5];
point[2] = a[2] < a[6] ? a[2] : a[6];
point[3] = a[3] < a[7] ? a[3] : a[7];
if (point[0] > point[2] || point[1] > point[3]) printf("0.00\n");
else printf("%.2lf\n", (point[2] - point[0])*(point[3] - point[1]));
}
system("pause");
return 0;
}