易知最短路一定是以圆心或者两圆交点作为中间点到达的。所以把这些点拿出来建图跑最短路就够了。

现在的问题就是，给定两个点，能否连边 add(a,b,dist(a,b))

题目要求，ab线段必须完全在圆上，所以可以求出ab线段和所有圆的所有交点，对于任意相邻两个交点，它们必处于同一个圆内，否则不可达。点的编号用map就够了（一开始我以为double有精度问题无法map，用两个longlong保存然后乘上1000000000，后来发现没问题，应该是这题都是整点，精度要求不高的原因吧）

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <map>
#include <cmath>
#include <cstring>
using namespace std;
#define pi acos(-1.0)
#define eps 1e-10
int cmp(double x)
{
    if(fabs(x)<eps) return 0;
    if(x>0) return 1;
    return -1;
}
double sqr(double x)
{
    return x*x;
}
struct point
{
    double x,y;
    point(){};
    point(double a,double b):x(a),y(b){};
    void out()
    {
        printf("%lf %lf\n",x,y);
    }
    void input()
    {
        scanf("%lf%lf",&x,&y);
    }
    friend point operator +(const point &a,const point &b)
    {
        return point(a.x+b.x,a.y+b.y);
    }
    friend point operator -(const point &a,const point &b)
    {
        return point(a.x-b.x,a.y-b.y);
    }
    friend bool operator ==(const point &a,const point &b)
    {
        return cmp(a.x-b.x)==0&&cmp(a.y-b.y)==0;
    }
    bool operator <(const point &a) const
    {
        if(x==a.x) return y<a.y;
        return x<a.x;
    }
    friend point operator *(const point &a,const double &b)
    {
        return point(a.x*b,a.y*b);
    }
    friend point operator *(const double &a,const point &b)
    {
        return point(a*b.x,a*b.y);
    }
    friend point operator /(const point &a,const double &b)
    {
        return point(a.x/b,a.y/b);
    }
    double norm()
    {
        return sqrt(sqr(x)+sqr(y));
    }
};
double det(const point &a,const point &b)
{
    return a.x*b.y-a.y*b.x;
}
double dot(const point&a,const point &b)
{
    return a.x*b.x+a.y*b.y;
}
double dist(const point &a,const point &b)
{
    return (a-b).norm();
}
point rotate_point(const point &p,double A)
{
    double tx=p.x,ty=p.y;
    return point(tx*cos(A)-ty*sin(A),tx*sin(A)+ty*cos(A));
}
point rotate(const point &p,double cost,double sint)
{
    double x=p.x,y=p.y;
    return point(x*cost-y*sint,x*sint+y*cost);
}

pair<point,point> crosspoint(point ap,double ar,point bp,double br)
{
    double d=(ap-bp).norm();
    double cost = (ar*ar + d*d -br*br)/(2*ar*d);
    double sint=sqrt(1.0-cost*cost);
    point v=(bp-ap)/(bp-ap).norm()*ar;
    return make_pair(ap+rotate(v,cost,-sint),ap+rotate(v,cost,sint));
}
void circle_cross_line(point a,point b,point o,double r,point ret[],int &num) {
    double x0 = o.x ,y0 = o.y;
    double x1 = a.x, y1 = a.y;
    double x2 = b.x, y2 = b.y;
    double dx = x2-x1, dy = y2-y1;
    double A = dx*dx+dy*dy;
    double B = 2*dx*(x1-x0) + 2*dy*(y1-y0);
    double C = sqr(x1-x0) + sqr(y1-y0)-sqr(r);
    double delta = B*B-4*A*C;
    if( cmp(delta) >= 0) {
        double t1 = (-B - sqrt(delta)) / (2*A);
        double t2 = (-B + sqrt(delta)) / (2*A);
        if(cmp(t1-1)<=0 && cmp(t1)>=0)
        ret[num++] = point(x1+t1*dx,y1+t1*dy);
        if(cmp(t2-1) <=0 && cmp(t2)>=0)
        ret[num++] = point(x1+t2*dx,y1+t2*dy);
    }
}
struct rad
{
    point c;
    double r;
}ra[300];
int n;
map<point,int> mp;
int ID;
int que[123456];
point xx[100005];
struct node2
{
    int v,next;
    double w;
}e[123456];
int head[12345];
bool in[12345];
double d[12345];
int en;
void add(int a,int b,double c)
{
  //  printf("%d %d %lf\n",a,b,c);
    e[en].v=b;
    e[en].w=c;
    e[en].next=head[a];
    head[a]=en++;

    e[en].v=a;
    e[en].w=c;
    e[en].next=head[b];
    head[b]=en++;
}
void spfa(int st)
{
    queue<int> q;
    memset(in,0,sizeof(in));
    for(int i=1;i<=ID;i++) d[i]=1000000000;
    q.push(st);
    in[st]=1;
    d[st]=0;
    int tmp;
    while(!q.empty())
    {
        tmp=q.front();q.pop();
        in[tmp]=0;
        for(int i=head[tmp];~i;i=e[i].next)
        {
            if(d[e[i].v]>d[tmp]+e[i].w)
            {
                d[e[i].v]=d[tmp]+e[i].w;
                if(!in[e[i].v])
                {
                    in[e[i].v]=1;
                    q.push(e[i].v);
                }
            }
        }
    }
}
bool inra(point &x,point &y)
{
    for(int i=1;i<=n;i++)
    {
        if(dist(x,ra[i].c)<=ra[i].r+eps && dist(y,ra[i].c)<=ra[i].r+eps)
        {
            return true;
        }
    }
    return false;
}
point my[12345];
bool cal(point &a,point &b)
{
    my[0]=a;
    int top=1;
    for(int i=1;i<=n;i++)
    {
        circle_cross_line(a,b,ra[i].c,ra[i].r,my,top);
    }
    my[top++]=b;
    sort(my,my+top);
    for(int i=1;i<top;i++)
    {
        if(!inra(my[i-1],my[i])) return false;
    }
    return true;
}
int main()
{
    int ca=1;
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        mp.clear();
        scanf("%d",&n);
        ID=0;
        for(int i=1;i<=n;i++)
        {
            ra[i].c.input();
            scanf("%lf",&ra[i].r);
        }
        printf("Case %d: ",ca++);

        if(cal(ra[1].c,ra[n].c))    //冲榜都靠这个特判#_#
        {
            printf("%.4f\n",dist(ra[1].c,ra[n].c));
            continue;
        }
        en=0;
        memset(head,-1,sizeof(head));
        int top=0;
        for(int i=1;i<=n;i++)
        {
            mp[ra[i].c]=++ID;
            que[top++]=ID;
            xx[ID]=ra[i].c;
            for(int j=i+1;j<=n;j++)
            {
                if(dist(ra[i].c,ra[j].c)>(ra[i].r+ra[j].r)+0.0) continue;
                pair<point,point> tmp=crosspoint(ra[i].c,ra[i].r,ra[j].c,ra[j].r);
                if(mp[tmp.first]==0)
                {
                    mp[tmp.first]=++ID;
                    que[top++]=ID;
                    xx[ID]=tmp.first;
                }
                if(mp[tmp.second]==0)
                {
                    mp[tmp.second]=++ID;
                    que[top++]=ID;
                    xx[ID]=tmp.second;
                }
            }
        }
        mp[ra[n].c]=++ID;
        que[top++]=ID;
        xx[ID]=ra[n].c;
        for(int i=0;i<top;i++)
        {
            for(int j=i+1;j<top;j++)
            {
                if(cal(xx[que[i]],xx[que[j]]))
                {
                    add(que[i],que[j],dist(xx[que[i]],xx[que[j]]));
                }
            }
        }
        spfa(1);

        if(d[ID]>=1000000000) puts("No such path.");
        else printf("%.4f\n",d[ID]);
    }
    return 0;
}
/*
99
2
0 0 1
2 0 1
2
0 0 1
4 1 2
4
-4 0 1
-1 0 2
1 0 2
4 0 1
3
-3 0 1
0 0 2
0 3 1
3
-3 0 1
0 0 2
-2 -1 1
3
-3 0 1
-2 -1 1
0 0 2
4
2 2 2
2 -2 2
-2 2 2
-2 -2 2
2
0 0 3
1 0 1
3
0 0 1
2 2 2
3 0 1
3
0 0 1
3 0 1
2 2 2

ans:
2.0000
No
8.0000
4.8284
1.4142
3.0000
6.8284
1.0000
3.0654
2.8284
*/

hdu 4063 Aircraft 计算几何+最短路