hdu 4454 Stealing a Cake 三分法

很容易想到三分法求解，不过要分别在0-pi，pi-2pi进行三分。

另外也可以直接暴力枚举……

代码如下：

 #include<iostream>
 #include<stdio.h>
 #include<algorithm>
 #include<iomanip>
 #include<cmath>
 #include<cstring>
 #include<vector>
 #define ll __int64
 #define pi acos(-1.0)
 #define MAX 50000
 using namespace std;
 struct point
 {
     double x,y;
     point(double _x=,double _y=){
         x=_x;
         y=_y;
     }
     point operator-(point a){
         return point(x-a.x,y-a.y);
     }
     point operator+(point a){
         return point(x+a.x,y+a.y);
     }
     double operator*(point a){
         return (x*a.x+y*a.y);
     }
 }p1,p2,p;
 double x,y,r;
 double dis2(point a,point b)
 {
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }
 double dis(point a, point b, point c)
 {
     point ab = b - a;
     point ac = c - a;
     double f = ab*ac;
     if (f<) return dis2(c, a);//C1处的点
     double d = ab*ab;
     if ( f>d) return dis2(c, b);//C2处的点，d=f*cos（theta）
     f = f/d;
     ab.x*=f;ab.y*=f;
     point D = a + ab; // c在ab线段上的投影点
     return dis2(c, D);
 }
 double line(point a)
 {
     point b,c;
     b.x=min(p1.x,p2.x);
     b.y=max(p1.y,p2.y);
     c.x=max(p1.x,p2.x);
     c.y=min(p1.y,p2.y);
     double MIN1,MIN2;
     MIN1=min(dis(p2,c,a),dis(p1,c,a));
     MIN2=min(dis(p1,b,a),dis(p2,b,a));
     return min(MIN1,MIN2);
 }
 point get(double a)
 {
     return point(x+r*cos(a),y+r*sin(a));
 }
 double solve()
 {
     double r,l,mid,midmid,t1,t2,ans1,ans2;
     point m1,m2;
     r=pi;l=;
     while(r-l>=1e-){
         mid=(r+l)/;
         midmid=(mid+r)/;
         m1=get(mid);
         m2=get(midmid);
         t1=line(m1)+dis2(m1,p);
         t2=line(m2)+dis2(m2,p);
         if(t1>t2) l=mid;
         else r=midmid;
     }
     ans1=t1;
     r=*pi;l=pi;
     while(r-l>=1e-){
         mid=(r+l)/;
         midmid=(mid+r)/;
         m1=get(mid);
         m2=get(midmid);
         t1=line(m1)+dis2(m1,p);
         t2=line(m2)+dis2(m2,p);
         if(t1>t2) l=mid;
         else r=midmid;
     }
     ans2=t1;
     return min(ans1,ans2);
 }
 int main(){
     while(cin>>p.x>>p.y){
         if(fabs(p.x)<1e-&&fabs(p.y)<1e-) break;
         cin>>x>>y>>r>>p1.x>>p1.y>>p2.x>>p2.y;
         point temp;
         temp.x=p1.x;temp.y=p1.y;
         p1.x=min(p1.x,p2.x);p1.y=min(p1.y,p2.y);
         p2.x=max(temp.x,p2.x);p2.y=max(temp.y,p2.y);
         printf("%.2lf\n",solve());
     }
     return ;
 }