hdu 4454（三分）

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三分有两种写法，一种是平均三分，在区间里分成三等分。
另一种是取中点，再取中点与右端点的中点进行三分。
在这题里，第一种可以过，第二种不能过。

所以以后还是都用平均的三分板子吧

#include<bits/stdc++.h>using namespace std;#define clr(x,y) memset(x,y,sizeof x)const int maxn = 10 + 10;struct Node{double x,y;};#define PI acos(-1.0)#define eps 1e-8Node p,pr,p1,p2;double rr;double dis(Node p1,Node p2){    return sqrt( (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));}double get1(double x){    double t = sqrt(rr * rr - (x - pr.x) * (x - pr.x));    double t1 = t + rr,t2 = rr - t;    return t1;}double get2(double x){    double t = sqrt(rr * rr - (x - pr.x) * (x - pr.x));    double t1 = t + rr,t2 = rr - t;    return t2;}double L,R,High,Low,l,r,high,low;double fun(double angle){    double ret = 1e18;    double x = pr.x + rr * cos(angle),y = pr.y + rr * sin(angle);    ret = min(ret,dis((Node){x,y},(Node){l,low}));    ret = min(ret,dis((Node){x,y},(Node){l,high}));    ret = min(ret,dis((Node){x,y},(Node){r,low}));    ret = min(ret,dis((Node){x,y},(Node){r,high}));    if(x >= l && x <= r)ret = min(ret,fabs(y - high)),ret = min(ret,fabs(y - low));    if(y >= low && y <= high)ret = min(ret,fabs(x - l)),ret = min(ret,fabs(x - r));    return ret + dis((Node){x,y},p);}int main(){   while( ~ scanf("%lf%lf",&p.x,&p.y))   {       if(p.x == 0 && p.y == 0)break;       scanf("%lf%lf%lf%lf%lf%lf%lf",&pr.x,&pr.y,&rr,&p1.x,&p1.y,&p2.x,&p2.y);       l = p1.x,r = p2.x,low = p1.y,high = p2.y;       if(l > r)swap(l,r);if(low > high)swap(low,high);        L = pr.x - rr,R = pr.x + rr,Low = pr.y - rr,High = pr.y + rr;        double LL = 0,RR = 2 * PI;        double ans;        while((RR - LL) > eps)        {//            cout << LL << " " << RR << endl;            double t = (RR - LL)/3.0;            double mid = LL + t,mmid = LL + 2 * t;            if(fun(mid) < fun(mmid))                ans = fun(mid),RR = mmid;            else ans = fun(mmid),LL = mid;        }       printf("%.2f\n",ans);   }   return 0;}

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