HDOJ1221 计算几何入门题

最新推荐文章于 2019-07-30 21:42:24 发布

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本文介绍了一个计算几何问题，即如何判断一个矩形与一个圆是否相交。通过输入矩形对角线和圆心坐标及半径，算法能够准确判断两者边界是否相交，并提供了完整的C++代码实现。

Rectangle and Circle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3237 Accepted Submission(s): 851

Problem Description
Given a rectangle and a circle in the coordinate system(two edges of the rectangle are parallel with the X-axis, and the other two are parallel with the Y-axis), you have to tell if their borders intersect.

Note: we call them intersect even if they are just tangent. The circle is located by its centre and radius, and the rectangle is located by one of its diagonal.

Input
The first line of input is a positive integer P which indicates the number of test cases. Then P test cases follow. Each test cases consists of seven real numbers, they are X,Y,R,X1,Y1,X2,Y2. That means the centre of a circle is (X,Y) and the radius of the circle is R, and one of the rectangle’s diagonal is (X1,Y1)-(X2,Y2).

Output
For each test case, if the rectangle and the circle intersects, just output “YES” in a single line, or you should output “NO” in a single line.

Sample Input
2
1 1 1 1 2 4 3
1 1 1 1 3 4 4.5

Sample Output
YES
NO

Author
weigang Lee

Source
杭州电子科技大学第三届程序设计大赛

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第一次做计算几何题目，感觉代码量比之前都大很多。这题还算是最最最简单的了。。。

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

struct node{
   double x,y;
}cir,a,b,c,d;
double r;

double dis(node &a, node &b){
    return sqrt(pow(a.x-b.x,2) + pow(a.y-b.y,2));
}

int main(){
   std::ios::sync_with_stdio(false);
   int p;
   node tmp;
   cin >> p;
   while (p--) {
       cin >> cir.x >> cir.y >> r >> a.x >> a.y >> b.x >> b.y;
       if (a.x>b.x) {tmp = a; a=b; b=tmp;}
       c.x = a.x; c.y = b.y;
       d.x = b.x; d.y = a.y;

       //这个最优先考虑，因为有可能圆在矩形内部，先判断后两种情况会输出YES，实际却是NO
       if (dis(a,cir)<r && dis(b,cir)<r && dis(c,cir)<r && dis(d,cir)<r){
           cout << "NO" << endl;
           continue;
       }

       if (a.x<=cir.x && b.x>=cir.x) {
          if (fabs(a.y-cir.y)<=r || fabs(b.y-cir.y)<=r) {
               cout <<"YES" << endl;
               continue;
          }
       }

       if ((a.y<=cir.y && b.y>=cir.y) ||(a.y>=cir.y && b.y<=cir.y)) {
         if (fabs(a.x-cir.x)<=r || fabs(b.x-cir.x)<=r) {
               cout <<"YES" << endl;
               continue;
          }
       }


       if (dis(a,cir)<=r || dis(b,cir)<=r || dis(c,cir)<=r || dis(d,cir)<=r) {
            cout <<"YES" << endl;
            continue;
       }

       cout << "NO" << endl;
   }
   return 0;
}