杭电2056之Rectangles 杭电

最新推荐文章于 2022-12-19 14:27:31 发布

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本文介绍了一种计算两个矩形交集面积的方法，输入为两对矩形对角线上的点坐标，输出为交集面积，精确到小数点后两位。文中提供了一个使用 C++ 实现的示例代码。

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

AC代码如下：
#include "iostream"
#include "algorithm"
#include "cstdio"
#include "cmath"
using namespace std;

int main(int argc, char* argv[])
{
    double m,n,area;
    double x1,x2,x3,x4,y1,y2,y3,y4,a[4],b[4];
    while(cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4)
    {
        a[0]=x1;a[1]=x2;a[2]=x3;a[3]=x4;
        b[0]=y1;b[1]=y2;b[2]=y3;b[3]=y4;
        sort(a,a+4);
        sort(b,b+4);
        m=fabs(x1-x2)+fabs(x3-x4)-(a[3]-a[0]);
        n=fabs(y1-y2)+fabs(y3-y4)-(b[3]-b[0]);
        area=m*n;
        if (m<0||n<0)
        {
            area=0;
        }
        printf("%.2lf\n",area);
    }
    return 0;
}