2017 Multi-University Training Contest
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6097
有一个圆,圆心在(0,0),有两点P,Q,不在圆外,且OP = OQ,求圆上一点D,使得DP+DQ最小,求这个最小的距离。
这个点不总是PQ垂直平分线与圆的交点,比赛的时候一开始我们也是这样想的,后来发现,当P,Q点离圆的边界很进的
时候,发现PQ所在线的与圆的焦点这条线就比垂直平分线那个焦点求出的距离要短,不过后来想的思路还是错了,最后
也没有过。
下面给两个博客的链接:
这个博客配图非常好:http://blog.csdn.net/qq_34845082/article/details/77099332
一看图就能明白了。
这个博客思路说得非常好:http://blog.csdn.net/merry_hj/article/details/77151650
至于题解中的反演点:百度百科查一下吧。这个题目就是一个数学题,败在不知道数学结论。
#include <iostream>#include <stdio.h>#include <math.h>using namespace std;///计算两点之间的距离double dist(double x1,double y1,double x2,double y2){ return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));}int main(){ int t; double px,py,qx,qy,r; scanf("%d",&t); while(t--) { double ans; scanf("%lf%lf%lf%lf%lf",&r,&px,&py,&qx,&qy); double OP = dist(0,0,px,py); double OP1 = r*r/OP; double POQ = acos((px*qx+py*qy)/(OP*OP))/2; ///角POQ的一半 double d = OP1*cos(POQ); if(px==qx && py==qy) { ans = 2*(r-dist(0,0,px,py)); printf("%.7lf\n",ans); } else if(px*px+py*py == r*r) { ans = dist(px,py,qx,qy); printf("%.7lf\n",ans); } else { if(d <= r) { ans = sqrt(OP1*OP1-d*d)*2; } else { double temp1 = OP1*OP1-d*d; double temp2 = (d-r)*(d-r); ans = sqrt(temp1+temp2)*2; } ans = ans*OP/r; printf("%.7lf\n",ans); } } return 0;}
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