HDU 2056 Rectangles (求两个相交矩形面积)_已知矩形对角线端点坐标求两矩形相交-优快云博客

本文链接：https://blog.youkuaiyun.com/a_ppler/article/details/10065315

本文介绍了一种计算两个矩形交集面积的方法，通过给定矩形对角线上的两个点坐标，利用简单的数学算法来高效求解。示例输入输出展示了具体的计算过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目大意：就是分别给出两个矩形 左下 和 右上两个对角线的坐标。求两个矩形的阴影面积。
我写的代码很水，有个代码很好。下面参考下。
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

int main()
{
    double x[4],y[4];
    while (scanf("%lf%lf",&x[0],&y[0])!=-1){
        for (int i=1;i<=3;i++) scanf("%lf%lf",&x[i],&y[i]);
        if (max(x[0],x[1])<min(x[2],x[3]) || min(x[0],x[1])>max(x[2],x[3]) || max(y[0],y[1])<min(y[2],y[3]) || min(y[0],y[1])>max(y[2],y[3])) printf("0.00\n");         
            //判断 两个矩形是否相交
           else
           {
            //如果相交，直接求阴影面积,这点很巧妙、
               sort(x,x+4); sort(y,y+4);
               printf("%.2lf\n",abs(x[1]-x[2])*abs(y[1]-y[2]));
           }      
    }   
}