hdu4454 三分 求点到圆,然后在到矩形的最短路

题意:
      求点到圆,然后在到矩形的最短路.


思路:

      把圆切成两半,然后对于每一半这个答案都是凸性的,最后输出两半中小的那个就行了,其中有一点,就是求点到矩形的距离,点到矩形的距离就是点到矩形四条边距离中最小的哪一个,求的方法有很多,一开始想都没想直接敲了个三分(这样就是两重三分了)跑了890ms AC的,后来看了看人家的都是直接用模板过的,我也找了个模板,结果31ms AC,哎,没事真的多暂歇模板,只有好处没坏处是了..

#include<stdio.h>#include<math.h>#include<algorithm>#define eps 1e-3 double PI=acos(-1.0);using namespace std;typedef struct{   double x ,y;}NODE;typedef struct{   NODE A ,B;}EDGE;NODE node[10];EDGE edge[10];NODE S ,O;double diss1[10] ,diss2[10];double R;bool camp(NODE a ,NODE b){   return a.x < b.x || a.x == b.x && a.y < b.y;}double dis(NODE a ,NODE b){   double tmp = pow(a.x - b.x ,2.0) + pow(a.y - b.y ,2.0);   return sqrt(tmp);}double minn(double x ,double y){   return x < y ? x : y;}double abss(double x){   return  x > 0 ? x : -x;}   NODE getnode(double du){   NODE ans;   ans.x = O.x + R *cos(du);   ans.y = O.y + R * sin(du);   return ans;}   //**********************************struct Point{    double x,y;    Point(double xx=0,double yy=0):x(xx),y(yy){}    Point operator -(const Point p1)    {        return Point(x-p1.x,y-p1.y);    }    double operator ^(const Point p1)    {        return x*p1.x+y*p1.y;    }};double cross(Point a,Point b){    return a.x*b.y-a.y*b.x;}inline int sign(double d){    if(d>eps)return 1;    else if(d<-eps)return -1;    else return 0;}double dis(Point A ,Point B){   return sqrt(pow(A.x - B.x ,2.0) + pow(A.y - B.y ,2.0));}double ptoline(Point tp,Point st,Point ed)//求点到线段的距离{    double t1=dis(tp,st);    double t2=dis(tp,ed);    double ans=min(t1,t2);    if(sign((ed-st)^(tp-st))>=0 && sign((st-ed)^(tp-ed))>=0)//锐角    {        double h=fabs(cross(st-tp,ed-tp))/dis(st,ed);        ans=min(ans,h);    }    return ans;}//************************ double search3_2(double L ,double R){   double low ,up ,mid ,mmid;     NODE now;   low = L ,up = R;   while(1)   {      mid = (low + up) / 2;      now = getnode(mid);      Point A ,B ,C;      A.x = now.x ,A.y = now.y;      for(int i = 1 ;i <= 4 ;i ++)      {            B.x = edge[i].B.x ,B.y = edge[i].B.y;         C.x = edge[i].A.x ,C.y = edge[i].A.y;         diss1[i] = ptoline(A ,B ,C) + dis(S ,now);              }      sort(diss1 + 1 ,diss1 + 4 + 1);          mmid = (mid + up) / 2;      now = getnode(mmid);            A.x = now.x ,A.y = now.y;      for(int i = 1 ;i <= 4 ;i ++)      {           B.x = edge[i].B.x ,B.y = edge[i].B.y;         C.x = edge[i].A.x ,C.y = edge[i].A.y;         diss2[i] = ptoline(A ,B ,C) + dis(S ,now);      }          sort(diss2 + 1 ,diss2 + 4 + 1);         if(diss1[1] > diss2[1]) low = mid;      else up = mmid;      if(abss(low - up) < eps) break;   }   return diss1[1];}int main (){   while(~scanf("%lf %lf" ,&S.x ,&S.y))   {      if(S.x == 0 && S.y == 0) break;      scanf("%lf %lf %lf" ,&O.x ,&O.y ,&R);      scanf("%lf %lf %lf %lf" ,&node[1].x ,&node[1].y ,&node[2].x ,&node[2].y);      node[3].x = node[1].x ,node[3].y = node[2].y;      node[4].x = node[2].x ,node[4].y = node[1].y;      sort(node + 1 ,node + 4 + 1 ,camp);      edge[1].A.x = node[1].x ,edge[1].A.y = node[1].y;      edge[1].B.x = node[2].x ,edge[1].B.y = node[2].y;       edge[2].A.x = node[2].x ,edge[2].A.y = node[2].y;      edge[2].B.x = node[4].x ,edge[2].B.y = node[4].y;       edge[3].A.x = node[4].x ,edge[3].A.y = node[4].y;      edge[3].B.x = node[3].x ,edge[3].B.y = node[3].y;       edge[4].A.x = node[3].x ,edge[4].A.y = node[3].y;      edge[4].B.x = node[1].x ,edge[4].B.y = node[1].y;      printf("%.2lf\n" ,minn(search3_2(0 ,PI) ,search3_2(PI ,2*PI)));   }   return 0;}