UVA - 10674-Tangents

最新推荐文章于 2018-08-13 21:20:46 发布

转载最新推荐文章于 2018-08-13 21:20:46 发布 · 84 阅读

本文介绍了一种通过模拟方法来求解两个圆之间的所有公切线的算法实现。该算法能正确处理各种情况，包括没有公切线、有一条公切线以及有多条公切线的情形，并给出了完整的C++代码实现。

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题意：给出两个圆，求它们的公切线，并依照一定格式输出

做法：模拟

代码：

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const double eps=2e-5;
const double pi=acos(-1.0);
int dcmp(double x){return fabs(x)<eps?

0:x<0?-1:1;} struct dot { double x,y; dot(){} dot(double a,double b){x=a;y=b;} dot operator +(dot a){return dot(x+a.x,y+a.y);} dot operator -(dot a){return dot(x-a.x,y-a.y);} dot operator *(double a){return dot(x*a,y*a);} double operator *(dot a){return x*a.y-y*a.x;} dot operator /(double a){return dot(x/a,y/a);} double operator /(dot a){return x*a.x+y*a.y;} bool operator ==(dot a){return x==a.x&&y==a.y;} void in(){scanf("%f%f",&x,&y);} void out(){printf("%f %f\n",x,y);} dot norv(){return dot(-y,x);} dot univ(){double a=mod();return dot(x/a,y/a);} dot ro(double a){return dot(x*cos(a)-y*sin(a),x*sin(a)+y*cos(a));} double mod(){return sqrt(x*x+y*y);} double dis(dot a){return sqrt(pow(x-a.x,2)+pow(y-a.y,2));} }; typedef pair<dot,dot> LL; LL ans[10]; bool operator <(LL a,LL b) { return dcmp(a.first.x-b.first.x)!=0?dcmp(a.first.x-b.first.x)<0: dcmp(a.first.y-b.first.y)<0; } int work(dot a,double r1,dot b,double r2) { int cnt; double ang; if(dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0&&dcmp(r1-r2)==0) return -1; if(dcmp(a.dis(b)+min(r1,r2)-max(r1,r2))<0) return 0; if(dcmp(a.dis(b)+min(r1,r2)-max(r1,r2))==0) { if(dcmp(r1-r2)>0) ans[0].first=ans[0].second=a+(b-a).univ()*r1; else ans[0].first=ans[0].second=b+(a-b).univ()*r2; return 1; } ang=acos((r1-r2)/a.dis(b)); ans[0].first=a+(b-a).ro(ang).univ()*r1; ans[1].first=a+(b-a).ro(-ang).univ()*r1; ang=pi-ang; ans[0].second=b+(a-b).ro(-ang).univ()*r2; ans[1].second=b+(a-b).ro(ang).univ()*r2; cnt=2; if(dcmp(a.dis(b)-r1-r2)==0) { ans[2].first=ans[2].second=a+(b-a).univ()*r1; cnt=3; } if(dcmp(a.dis(b)-r1-r2)>0) { ang=acos((r1+r2)/a.dis(b)); ans[2].first=a+(b-a).ro(ang).univ()*r1; ans[3].first=a+(b-a).ro(-ang).univ()*r1; ans[2].second=b+(a-b).ro(ang).univ()*r2; ans[3].second=b+(a-b).ro(-ang).univ()*r2; cnt=4; } sort(ans,ans+cnt); return cnt; } int main() { int i,cnt; dot a,b; double r1,r2; while(cin>>a.x>>a.y>>r1>>b.x>>b.y>>r2) { if(a.x==0&&a.y==0&&r1==0&&b.x==0&&b.y==0&&r2==0) return 0; cnt=work(a,r1,b,r2); cout<<cnt<<endl; for(i=0;i<cnt;i++) printf("%.5f %.5f %.5f %.5f %.5f\n",ans[i].first.x,ans[i].first.y, ans[i].second.x,ans[i].second.y,ans[i].first.dis(ans[i].second)); } }

Description

Problem B
Tangents
Input: standard input
Output: standard output
Time Limit: 2 seconds

You can see in the pictures below that two different circles can have at most four common tangents. Given the center and radius of two circles your job is to find the length of their common tangents and also the points where they touch the two circles.

Input

The input file contains several lines of inputs.

Each line contains six integers x₁ (-100<=x₁<=100), y₁ (-100<=y₁<=100), r₁ (0<r₁<=200), x₂ (-100<=x₂<=100), y₂ (-100<=y₂<=100), r₂ (0<r₂<=200). Here (x₁, y₁) and (x₂, y₂) are the coordinates of the center of the first circle and second circle respectively, r1 is the radius of the first circle and r2 is the radius of the second circle.

Input is terminated by a line containing six zeroes.

Output

For each line of input you should produce one of more lines of output. The description of this output is given below.

First line of the output for each line of input contains an integer n, which denotes the number of different tangents between the two circles. If there is infinite number of tangents between the two circles then the value of n should be –1. If n is positive then next n lines contains the description of each tangent. The description of the tangent contains five floating-point numbers S_x, S_y, T_x, T_y, L in a single line. Here (S_x, S_y) is the point at which the tangent touches the first circle and (T_x, T_y) is the point where the tangent touches the second circle and L is the length of the tangent. All the floating-point numbers have five digits after the decimal point. Errors less than 2*10^-5 will be ignored. The tangents should be printed in ascending order of S_x and in case of a tie they should be printed in ascending order of S_y.