UCF Local Programming Contest 2018 Circle Meets Square （圆与正方形位置关系的判断）

We all know that you can’t put a round peg in a square hole. Asking you to do so in this contestwould be cruel and unusual punishment, which is banned by the Eighth Amendment to the UnitedStates Constitution. But, perhaps a more reasonable problem that the Framers of the Constitutionnever thought about is determining if a given circle and square have an intersection of positivearea (overlap), or touch (share a point in common), or don’t touch at all. 

The Framers of the US Constitution and the UCF Programming Team coaches would like to know,given a circle and a square, do the two overlap in some positive area, touch, or don’t touch at all.Help them out by writing a program to solve the problem! 

The Problem:

Given the description of a square and circle in the Cartesian plane, determine if the intersectionbetween the two has positive area (overlap), is a single point (touches) or doesn’t exist.

The Input: 

The first line of input contains three integers: x (-1,000 ≤ x ≤ 1,000), y (-1,000 ≤ x ≤ 1,000), and r(0 < r ≤ 1,000), representing the x and y coordinates and radius, respectively of the circle. 

The second line of input contains three integers: tx (-1,000 ≤ tx < 1,000), ty (-1,000 ≤ ty < 1,000),and s (0 < s ≤ 1,000), where (tx, ty) represents the coordinates of the bottom left corner of the squareand s represents the side length of the square. The square’s top right corner is (tx+s, ty+s), so thatits sides are parallel to the x and y axes. 

The Output: 

If the circle and square don’t touch, output 0 (zero). If they touch at a single point, output 1(one). If they overlap in positive area, output 2 (two).

题目大意与分析

本题是判断圆与正方形的关系，是相交、只有一个点接触还是相离

对于相交（包括包含）的判断，有两种情况：正方形四个顶点有至少一个在圆内 或者 圆的上下左右四个点与圆心这五个点至少有一个在正方形内

对于只有一个点接触，是在不是上述关系的基础上，如果正方形的四个顶点有一个在圆上，或者圆的上下左右四个点有一个点在正方形上

其余情况为相离

#include<bits/stdc++.h>

using namespace std;

int zx,zy,sx,sy,d,r;

int incirle(int x,int y)        //判断点在圆内 用点到圆心距离判断 
{
    if(((x-sx)*(x-sx)+(y-sy)*(y-sy))<(r*r))
    {
        return 1;
    }
    else
    {
        return 0;
    }
} 

int oncirle(int x,int y)        //判断点在圆上 用点到圆心距离判断 
{
    if(((x-sx)*(x-sx)+(y-sy)*(y-sy))==(r*r))
    {
        return 1;
    }
    else
    {
        return 0;
    }
} 

int insquare(int x,int y)     //判断点在正方形内 用坐标关系判断 
{
    if((x>zx)&&(x<zx+d)&&(y>zy)&&(y<zy+d))
    {
        return 1;
    }
    else
    {
        return 0;
    }
}

int onsquare(int x,int y)     //判断点在正方形上 用坐标关系判断 
{
    if(((x==zx)&&(y>=zy)&&(y<=zy+d))||((x==zx+d)&&(y>=zy)&&(y<=zy+d))||((y==zy)&&(x>=zx)&&(x<=zx+d))||((y==zy+d)&&(x>=zx)&&(x<=zx+d)))
    {
        return 1;
    }
    else
    {
        return 0;
    }
}

int pd2()
{
    if(incirle(zx,zy)||incirle(zx+d,zy)||incirle(zx,zy+d)||incirle(zx+d,zy+d))  //某个顶点在圆内 
    {
        return 1;
    }
    if(insquare(sx,sy)||insquare(sx+r,sy)||insquare(sx-r,sy)||insquare(sx,sy+r)||insquare(sx,sy-r)) //圆的点在正方形内 
    {
        return 1;
    }
    else
    {
        return 0;
    }
}


int pd1()
{
    if(oncirle(zx,zy)||oncirle(zx+d,zy)||oncirle(zx,zy+d)||oncirle(zx+d,zy+d))   //点在圆上
    {
        return 1;
    }
    if(onsquare(sx+r,sy)||onsquare(sx-r,sy)||onsquare(sx,sy+r)||onsquare(sx,sy-r))  //点在正方形上 
    {
        return 1; 
    }
    else
    {
        return 0;
    }
}


int main()
{
    cin>>sx>>sy>>r;
    cin>>zx>>zy>>d;
    if(pd2())
    cout<<"2"<<endl;
    else if(pd1())
    cout<<"1"<<endl;
    else
    cout<<"0"<<endl;
 } 

原文地址：https://www.cnblogs.com/dyhaohaoxuexi/p/12575238.html