题目描述:

在二维坐标系中,所有的值是double类型,那么一个矩形可以由四个点来代表,(x1, y1)为最左的点,(x2, y2)为最上的点,(x3, y3)为最下的点,(x4, y4)为最右的点。
给定4个点代表的矩形,再给定一个点(x, y),判断(x, y)是否在矩形中

题目分析:

  1. 矩形的边平行于坐标轴
此时(x1, y1)为左上角的点,(x2, y2)为右上角的点,(x3, y3)为左下角的点,(x4, y4)为右下角的点。这种情况很好判断。

  1. 矩形的边不平行于坐标轴
此时,需要旋转矩形,使之平行于坐标轴。

image

#include<iostream>
#include<cmath>
using namespace std;

bool isInside(double x1, double x4, double y1, double y4, double x, double y)
{
    if(x <= x1) return false;
    if(x >= x4) return false;
    if(y >= y1) return false;
    if(y <= y4) return false;
    return true;
}

bool isInside(double x1, double y1, double x2, double y2, double x3, double y3,
             double x4, double y4, double x, double y)
{
    if(x1 == x3) return isInside(x1, x4, y1, y4, x, y);
    double l = y4 - y3;
    double k = x4 - x3;
    double s = sqrt(k * k + l * l);
    double sin = l / s;
    double cos = k / s;
    double x1r = cos * x1 + sin * y1;
    double y1r = -x1 * sin + y1 * cos;
    double x4r = cos * x4 + sin * y4;
    double y4r = -x4 * sin + y4 * cos;
    double xr = cos * x + sin * y;
    double yr = -x * sin + y * cos;
    return isInside(x1r, x4r, y1r, y4r, xr, yr);
}
int main()
{
    double x1, y1, x2, y2, x3, y3, x4, y4, x, y;
    cin >> x1 >> y1;
    cin >> x2 >> y2;
    cin >> x3 >> y3;
    cin >> x4 >> y4;
    cin >> x >> y;
    if(isInside(x1, y1, x2, y2, x3, y3, x4, y4, x, y)) 
    {
        cout << "Yes" << endl;
    }
    else 
    {
        cout << "No" <<endl;
    }
    return 0;
}
posted on 2020-01-21 22:53  xiaobaizzZ  阅读(6418)  评论(0编辑  收藏  举报