HDU 4454 Stealing a Cake --枚举

题意： 给一个点，一个圆，一个矩形， 求一条折线，从点出发，到圆，再到矩形的最短距离。

解法： 因为答案要求输出两位小数即可，精确度要求不是很高，于是可以试着爆一发，暴力角度得到圆上的点，然后求个距离，求点到矩形的距离就是求点到四边的距离然后求个最小值，然后总的取个最小值即可。

代码：

#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm>

#define Mod 1000000007

#define pi acos(-1.0)

#define eps 1e-8

using namespace std; #define N 1000107 typedef struct Point { double x,y; Point(double x=0,double y=0):x(x),y(y){} Point(){} }Vector; int sgn(double x) { if(x > eps) return 1; if(x < -eps) return -1; return 0; } Vector operator + (Vector A,Vector B) { return Vector(A.x+B.x,A.y+B.y); } Vector operator - (Point A,Point B) { return Vector(A.x-B.x,A.y-B.y); } bool operator == (const Point& a,const Point& b) { return (sgn(a.x-b.x) == 0 && sgn(a.y-b.y) == 0); } double Dot(Vector A,Vector B) { return A.x*B.x + A.y*B.y; } double Cross(Vector A,Vector B) { return A.x*B.y - A.y*B.x; } double Length(Vector A) { return sqrt(Dot(A,A)); } double PtoSeg(Point P,Point A,Point B) { if(A == B) return Length(P-A); Vector V1 = B-A; Vector V2 = P-A; Vector V3 = P-B; if(sgn(Dot(V1,V2)) < 0) return Length(V2); else if(sgn(Dot(V1,V3)) > 0) return Length(V3); else return fabs(Cross(V1,V2))/Length(V1); } int main() { double X,Y; double Cx,Cy,R; double x1,y1,x2,y2; while(scanf("%lf%lf",&X,&Y)!=EOF) { if(sgn(X) == 0 && sgn(Y) == 0) break; scanf("%lf%lf%lf",&Cx,&Cy,&R); scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); Point A = Point(x1,y1); Point B = Point(x1,y2); Point C = Point(x2,y1); Point D = Point(x2,y2); double delta = 2.0*pi*0.0001; double Min = Mod; for(int Angle=1;Angle<=10000;Angle++) { double ang = delta*Angle; double nx = Cx + R*cos(ang); double ny = Cy + R*sin(ang); double D1 = PtoSeg(Point(nx,ny),A,B); double D2 = PtoSeg(Point(nx,ny),A,C); double D3 = PtoSeg(Point(nx,ny),B,D); double D4 = PtoSeg(Point(nx,ny),C,D); double Dis = sqrt((nx-X)*(nx-X)+(ny-Y)*(ny-Y))+min(min(min(D1,D2),D3),D4); Min = min(Dis,Min); } printf("%.2f\n",Min); } return 0; }

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