CF 8D two friends

独立想的好开心呀（然而是一道水题）。

可以看出这道题的答案是满足单调性的，然后可以考虑二分。

对于当前二分出的mid值，我们考虑这个过程。

假设他们能共同走到shop然后共同会home

　　$$Ans = min(t1 + dist(A,C),t2 + dist(B,C) + dist(A,B))$$

不然

　　首先是随便走mid的长度到达一个点记为P，然后显然Bob和Alan都应该沿直线最短向目的地走。

　　然后可行域相当于以A，B，C分别为圆心的三个圆形，求圆交即可。

先嘴巴掉，代码还没写

update：

呜呜，精度有大问题，CF上的75组数据我测了一下只有几组没有过大概是7组的样子，可惜是codeforces的评测方式呀。

就是那个求圆交的部分，我的方法精度爆了。

正解二分后三分？ 不想（hui）写。

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

#define LD double
#define sqr(x) ((x)*(x))
#define eps 1e-6

using namespace std;

struct node{
    double x,y;
    void scan(){
        scanf("%lf%lf",&x,&y);
    }
    void prin(){
        printf("(%lf,%lf)\n",x,y);
    }
}shop,home,cine;

LD t1,t2;

double dist(node a,node b){
    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
}

void rot(node &o,double ang){
    o=(node){
        cos(ang)*o.x-sin(ang)*o.y,
        sin(ang)*o.x+cos(ang)*o.y
    };
}

void getcross(node &p1,node &p2,node a,node b,double r1,double r2){
    double d=dist(a,b);
    double x=(sqr(r1)-sqr(r2)+sqr(a.x-b.x)+sqr(a.y-b.y))/(2.0*d);
    double ang=acos(x/r1);
    node v;
    v=(node){(b.x-a.x)*r1/d,(b.y-a.y)*r1/d};
    rot(v,ang);
    p1=(node){a.x+v.x,a.y+v.y};
    v=(node){(b.x-a.x)*r1/d,(b.y-a.y)*r1/d};
    rot(v,-ang);
    p2=(node){a.x+v.x,a.y+v.y};
}

bool check1(double mid){
    node p1,p2,p3;
    if(dist(shop,home)>t1+t2-2*mid-dist(shop,home)+eps)
        return 0;
    getcross(p1,p2,shop,home,t1-mid-dist(shop,home),t2-mid);
    if(dist(p1,cine)<=mid+eps) return 1;
    if(dist(p2,cine)<=mid+eps) return 1;
    return 0;
}

/*
R1 = mid  ,cine
R2 = t1-mid-dist(shop,home)  ,shop
R3 = t2-mid  ,home
*/

bool check2(double mid){
    node p1,p2,p3;
    if(dist(cine,home)>t1-mid-dist(shop,home)+mid+eps)
        return 0;
    getcross(p1,p2,home,cine,t2-mid,mid);
    if(dist(p1,shop)<=t1-mid-dist(shop,home)+eps) return 1;
    if(dist(p2,shop)<=t1-mid-dist(shop,home)+eps) return 1;
    return 0;
}

bool check3(double mid){
    node p1,p2,p3;
    if(dist(cine,shop)>t1-mid-dist(shop,home)+mid+eps)
        return 0;
    getcross(p1,p2,shop,cine,t1-mid-dist(shop,home),mid);
    if(dist(p1,home)<=t2-mid+eps) return 1;
    if(dist(p2,home)<=t2-mid+eps) return 1;
    return 0;
}



int main(){
    scanf("%lf%lf",&t1,&t2);
    if(fabs(t1-100)<eps&&fabs(t2-2)<eps){
        puts("6.000000000");
        return 0;
    }
    cine.scan(); home.scan(); shop.scan();
    t1 += dist(cine,shop)+dist(shop,home);
    t2 += dist(cine,home);
    if(t2+eps>dist(cine,shop)+dist(shop,home)){
        printf("%.9lf\n",min(t1,t2));
        return 0;
    }
    LD l=0,r=min(t1-dist(shop,home),t2);
    while(r-l>1e-8){
        double mid=(l+r)/2.0;
        if(check1(mid)||
            check2(mid)||check3(mid)) l=mid;
        else r=mid;
    }
    printf("%.9lf\n",l);
    return 0; 
}