标签:else long space ace time cst fir debug back
题意:在平面中,有一个圆,有一个点,问能在这个圆中围出最大的圆的圆心坐标和半径。要求这个最大圆不包含这个点。
思路:比较基础的计算几何,要分三种情况,第一种就是这个点在圆外的情况。第二种是点在圆内。第三种是这个点和圆心重合。
ac代码
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <list> #include <iterator> #include <cmath> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue #define Pll pair<ll,ll> #define Pii pair<int,int> #define fi first #define se second #define OKC ios::sync_with_stdio(false);cin.tie(0);cout.tie(0) typedef long long ll; typedef unsigned long long ull; const double espp = 0.0000000001; /*-----------------show time----------------*/ double R,x,y,x2,y2; double dis(double a,double b,double x,double y) { return sqrt((a-x)*(a-x) + (b-y)*(b-y)); } int main(){ scanf("%lf%lf%lf%lf%lf", &R, &x, &y, &x2, &y2); double d = dis(x,y,x2,y2); if(d > R ||abs(d-R) < espp) { printf("%lf %lf %lf\n",x,y,R); } else { double r1 = (R + d)/2; double dx = 1.0*(x-x2)*(x-x2) + 1.0*(y-y2)*(y-y2); dx = sqrt(dx); double ans1,ans2; if(dx!=0) { ans1 = x2 + (x-x2)/dx*r1; ans2 = y2 + (y-y2)/dx*r1; } else { ans1 = x; ans2 = y+R/2; } printf("%.9lf %.9lf %.9lf\n",ans1,ans2,r1); } return 0; }
Codeforces Round #465 &935C. Fifa and Fafa计算几何
标签:else long space ace time cst fir debug back
原文地址:https://www.cnblogs.com/ckxkexing/p/9135013.html