B - Space Bowling
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87643#problem/B
Description
The inhabitants of planets orbiting around the pulsar PSR 2010+15 enjoy playing space bowling. A few cylindrical pins of unit diameter are set on a huge field. A player chooses a certain point of the field and rolls a ball from this point, trying to destroy as many pins as possible. After the ball is released, it rolls in a straight line, touching the surface all the time before rolling away from the field. If the ball touches a pin, this pin dematerializes, and the ball doesn't change direction. To score a strike, the player has to destroy at least k pins in one shot.
Unfortunately, aliens haven't yet invented a machine that would return the balls that rolled away from the field. Instead, they use a machine that materializes a new ball from vacuum before each shot. A player enters the diameter and in a second he obtains a ball of exactly the same diameter.
It is time for an alien Vas-Vas to roll a ball. There are n pins standing on the field at the moment. Help Vas-Vas to determine the minimal diameter of a ball, he can score a strike with.
Input
The first line contains space-separated integers n and k (1 ≤ k ≤ n ≤ 200) . The i-th of following n lines contains space-separated integersxi and yi (−10 5 ≤ xi, yi ≤ 10 5) , which are the coordinates of the centers of pins. All pins are situated at different points.
Output
Output the minimal possible diameter of a ball which can be used to score a strike, with absolute or relative error not exceeding 10 −6. If a strike can be scored with a ball of arbitrarily small diameter, output “0.000000”.
Sample Input
5 4
0 4
0 6
6 4
6 6
3 0
Sample Output
1.0000000000
HINT
题意
平面上有n个直径为1的圆，然后让你飞出一个球，要求触碰至少k个圆，问这个球的半径至少为多少
题解：
直接暴力枚举两个圆，然后直线方向就是这两个圆的中垂线，然后暴力算距离，取第k大的就好了
有一些小细节
代码前面都是模板，无视就好了
代码：
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <complex> #include <vector> using namespace std; const double eps = 1e-10; //精度控制 const double pi = acos(-1); int dcmp(double p) // 三态比较, 负数返回 -1 ， 0 返回 0 ， 正数返回 1 { if (fabs(p) < eps) return 0; else return p < 0 ? -1 : 1; } inline double Deg2Rad(const double & p) // 角度转弧度 { return pi*p/180.0; } inline double Rad2Deg(const double & p) // 弧度转角度 { return p * 180.0 / pi; } struct Point { long long x , y ; Point (double x = 0. , double y = 0.) { this->x = x , this->y = y; } //***********重载函数***************** friend Point operator + (const Point & T1, const Point & T2) { return Point(T1.x + T2.x , T1.y + T2.y); } friend Point operator - (const Point & T1, const Point & T2) { return Point(T1.x - T2.x , T1.y - T2.y); } friend Point operator * (const Point & T1 , const double p) { return Point(T1.x * p , T1.y * p); } friend Point operator * (const Point & T1 , const int p) { return Point(T1.x * p , T1.y * p); } friend bool operator == (const Point & T1 , const Point & T2) // 三态比较，精度较高 { return dcmp(T1.x - T2.x) == 0 && dcmp(T1.y - T2.y) == 0; } friend ostream& operator << (ostream & os,const Point & x) { os << "x is " << x.x << " y is " << x.y; return os; } //************************************ }; typedef Point Vector; double Dot(const Vector & T1 , const Vector & T2) { return T1.x * T2.x + T1.y * T2.y; } double Length(const Vector & T1) { return sqrt(Dot(T1,T1)); } double Angle(const Vector & T1 , const Vector & T2) { return acos(Dot(T1,T2)/ Length(T1) / Length(T2)); } double Cross(const Vector & T1, const Vector & T2) { return T1.x * T2.y - T2.x * T1.y; } double Area2(const Point & T1 , const Point & T2, const Point & T3) { return Cross(T3-T1,T2-T1); } // 将T1向量绕着起点逆时针旋转 rad 弧度 Vector Rotate(const Vector & T1 , const double rad) { return Vector(T1.x*cos(rad) - T1.y * sin(rad) , T1.x*sin(rad) + T1.y*cos(rad)); } // 返回向量T1的单位向量,调用时确保向量长度不为 0 Vector Normal(const Vector & T1) { double L = Length(T1); return Vector(T1.x/L,T1.y/L); } bool cmp1(const Point & a,const Point & b) { return a.x < b.x || (dcmp(a.x-b.x) == 0 && a.y < b.y); } //按照极角排序 bool cmp2(const Point & a,const Point & b) { double t1 = atan2(a.y,a.x); double t2 = atan2(b.y,b.x); if (t1 < 0) t1 += 360.0; if (t2 < 0) t2 += 360.0; return t1 < t2; } // 点 A 到 点 u 和点 v 组成的直线的距离 double DistanceToLine(const Point & A , const Point & u , const Point &v) { Vector v1(A-u) , v2(v-u); return fabs(Cross(v1,v2)) / Length(v2); } double DistanceToSegment(const Point & A,const Point & u,const Point & v) { if (u == v) return Length(A-u); Vector v1 = v - u , v2 = A - u , v3 = A - v; if (dcmp(Dot(v1,v2)) < 0) return Length(v2); else if(dcmp(Dot(v1,v3)) > 0) return Length(v3); else return fabs(Cross(v1,v2)) / Length(v1); } // 求两直线交点 , Line A上一点a , 向量 va , Line B上一点 b ,向量vb Point GetLineIntersection(const Point & a , const Vector & va, const Point & b , const Vector & vb) { Vector u = a-b; double t = Cross(vb, u) / Cross(va, vb); return a+va*t; } // 规范相交判断 bool SegmentProperIntersection(const Point & a1,const Point & a2,const Point & b1,const Point & b2) { double c1 = Cross(b2-b1,a1-b1) , c2 = Cross(b2-b1,a2-b1) , c3 = Cross(a2-a1,b1-a1) , c4 = Cross(a2-a1,b2-a1); return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0; } // 判断点p是否在 a1 和 a2构成的线段上( 不包括端点 ) bool IsPointOnSegment(const Point & p,const Point & a1 , const Point & a2) { return dcmp(Cross(a1-p,a2-p)) == 0 && Dot(a1-p,a2-p) < 0; } // Pay attention !!! // 下面的多边形函数带入的多边形点集必须全部为多边形上的点 ， 且 必须已经按照极坐标排序(逆时针) // 返回凸包的面积，p 必须全部是凸包上的点 double ConvexPolygonArea(Point * Polygon , int n) { double area = 0; for(int i = 1 ; i < n - 1 ; ++ i) area += Cross(Polygon[i]-Polygon[0],Polygon[i+1]-Polygon[0]); return area / 2.0; } // 复杂度 O(n) // 点在多边形内的判断 int PointInPolygon(const Point & p,const Point * Polygon,const int n) { int wn = 0; for(int i = 0 ; i < n ; ++ i) { int t1 = i; int t2 = (i+1) >= n ? i+1-n : i+1; if (IsPointOnSegment(p,Polygon[t1],Polygon[t2])) return -1; // 点在多边形边界上 int k = dcmp(Cross(Polygon[t2] - Polygon[t1],p - Polygon[t1])); int d1 = dcmp(Polygon[t1].y - p.y); int d2 = dcmp(Polygon[t2].y - Polygon[t1].y); if (k > 0 && d1 <= 0 && d2 > 0) wn ++; if (k < 0 && d2 <= 0 && d1 > 0) wn --; } if (wn != 0) return 1; //在多边形内部 return 0; // 外部 } //********************************************************************************************** // 计算凸包 // 输入的p数组不允许有重复点 // 如果不希望在凸包的边上有输入点，把两个<= 该成 < // 精度要求较高时用dcmp int ConvexHull(Point * p , int n , Point * ch) { sort(p,p+n,cmp1); int m = 0; for(int i = 0 ; i < n ; ++ i) { while(m > 1 && Cross(ch[m-1] - ch[m-2] , p[i] - ch[m-2]) <= 0 ) m --; ch[m++] = p[i]; } int k = m; for(int i = n - 2 ; i >= 0 ; -- i) { while(m > k && Cross(ch[m-1] - ch[m-2] , p[i] - ch[m-2]) <= 0 ) m --; ch[m++] = p[i]; } if (n > 1) m--; return m; } const int maxn = 2e2 + 50; int n , k ,cot; Point p[maxn]; long long pos[maxn]; int main(int argc,char *argv[]) { double ans = 1e200; scanf("%d%d",&n,&k); if (n == 1) { printf("0 "); return 0; } for(int i = 0 ; i < n ; ++ i) scanf("%I64d%I64d",&p[i].x,&p[i].y); for(int i = 0 ; i < n ; ++ i) for(int j = 0; j < n ; ++ j) { // if (i == j) continue; long long dx = p[j].y - p[i].y; long long dy = p[i].x - p[j].x; long long c = dx * p[i].x + dy * p[i].y; int cur = 0; for(int kk = 0 ; kk < n ; ++ kk) { long long d = dx * p[kk].x + dy*p[kk].y; if (d - c >= 0) { pos[cur++] = d-c; } } if (cur >= k) { sort(pos,pos+cur); ans = min(ans , (double)pos[k-1] / sqrt(dx*dx + dy*dy)); } } if (dcmp(ans-1.00) <= 0) ans = 0.0; else ans -= 1.00; printf("%.10lf ",ans); return 0; }