C - Rectangles

最新推荐文章于 2019-02-03 18:42:03 发布

原创最新推荐文章于 2019-02-03 18:42:03 发布 · 490 阅读

CC 4.0 BY-SA版权

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顶点坐标，输出精确到小数点后两位的交集面积。

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.

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


本题的解法过程在网上都可以查到，但本题的坑点是，必须持续输入，直到输入出错停止

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;



int main()
{
    double x1,y1,x2,y2,x3,y3,x4,y4;
    while(cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4){
    if(x1>x2)swap(x1,x2);
    if(y1>y2)swap(y1,y2);
    if(x3>x4)swap(x3,x4);
    if(y3>y4)swap(y3,y4);
    double X1,X2,Y1,Y2;
    X1=max(x1,x3);X2=min(x2,x4);
    Y1=max(y1,y3);Y2=min(y2,y4);
    if(X1>=X2||Y1>=Y2)printf("%.2lf\n",0);
    else printf("%.2lf\n",(X2-X1)*(Y2-Y1));
    }
    return 0;
}