不说了。。。说多了都是泪。。。从昨天下午一直wa到现在，直到刚刚才让人帮我找到所谓的“bug”，其实也算不上bug。。。

这个题的思路就是：找出平面上的所有点：所有圆的交点以及所有圆的圆心。然后依次判断两点是否连通，连通的话两点距离便是其欧几里得距离。这样建完图之后直接跑s->t最短路就行了。。

两点连通？也就是说这两点连成的线段，一直在圆内，任意圆都行。如何判断呢，求出该线段与所有圆的所有交点，排序后将其分段，依次判断每一段是否在任意圆内。这个么，在分段后，判断每一段的中点是否在圆内就行了。

这题并不是很难啊？我的bug在哪？？刚开始我是将所有圆的圆心都加到平面点集中，wa到死有木有。。。刚刚改了改，只加第一个跟最后一个圆的圆心。。立A。。为什么？

#include<algorithm>

#include<iostream>

#include<cstring>

#include<fstream>

#include<sstream>

#include<vector>

#include<string>

#include<cstdio>

#include<bitset>

#include<queue>

#include<stack>

#include<cmath>

#include<map>

#include<set>

#define FF(i, a, b) for(int i=a; i<b; i++)

#define FD(i, a, b) for(int i=a; i>=b; i--)

#define REP(i, n) for(int i=0; i<n; i++)

#define CLR(a, b) memset(a, b, sizeof(a))

#define LL long long

#define PB push_back

#define eps 1e-10

#define debug puts("**debug**")

using namespace std;

const double PI = acos(-1);

const double INF = 1e20;

const int maxn = 11111;

struct Point

{

    double x, y;

    Point (double x=0, double y=0):x(x), y(y) {}

};

typedef Point Vector;

struct Circle

{

   Point c;

   double r;

   Circle() {}

   Circle(Point c, double r) : c(c), r(r) {}

   Point point(double a) { return Point(c.x+cos(a)*r, c.y+sin(a)*r); }

};

struct Line

{

    Point p;

    Vector v;

    double ang;

    Line(){}

    Line(Point p, Vector v) : p(p), v(v) {ang = atan2(v.y, v.x); }

    Point point(double t)

    {

        return Point(p.x + t*v.x, p.y + t*v.y);

    }

    bool operator < (const Line& L) const

    {

        return ang < L.ang;

    }

};

template <class T> T sqr(T x) { return x * x;}

Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }

Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }

Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }

Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }

bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }

bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }

int dcmp(double x)

{

    if(fabs(x) < eps) return 0;

    return x < 0 ? -1 : 1;

}

bool operator == (const Point& a, const Point& b){ return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }

double Length(Vector A) { return sqrt(Dot(A, A)); }

double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }

Vector vecunit(Vector x){ return x / Length(x);}

Vector normal(Vector x) { return Point(-x.y, x.x) / Length(x);}

double angel(Vector v) { return atan2(v.y, v.x); }

bool OnSegment(Point p, Point a1, Point a2)

{

    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;

}

bool InCircle(Point x, Circle c) { return dcmp(sqr(c.r) - sqr(Length(c.c - x))) > 0;}

double DistanceToSeg(Point p, Point a, Point b)

{

    if(a == b) return Length(p-a);

    Vector v1 = b-a, v2 = p-a, v3 = p-b;

    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);

    if(dcmp(Dot(v1, v3)) > 0) return Length(v3);

    return fabs(Cross(v1, v2)) / Length(v1);

}

void getLineABC(Point A, Point B, double& a, double& b, double& c)

{

    a = A.y-B.y, b = B.x-A.x, c = A.x*B.y-A.y*B.x;

}

Point GetIntersection(Line a, Line b)

{

    Vector u = a.p-b.p;

    double t = Cross(b.v, u) / Cross(a.v, b.v);

    return a.p + a.v*t;

}

int getCCintersection(Circle C1, Circle C2, vector<Point>& sol)

{

    double d = Length(C1.c - C2.c);

    if(dcmp(d) == 0)

    {

        if(dcmp(C1.r - C2.r) == 0) return -1;

        return 0;

    }

    if(dcmp(C1.r + C2.r - d) < 0) return 0;

    if(dcmp(fabs(C1.r - C2.r) - d) > 0) return 0;

    double a = angel(C2.c - C1.c);

    double da = acos((sqr(C1.r) + sqr(d) - sqr(C2.r)) / (2*C1.r*d));

    Point p1 = C1.point(a-da), p2 = C1.point(a+da);

    sol.PB(p1);

    if(p1 == p2) return 1;

    sol.PB(p2);

    return 2;

}

int getSegCircleIntersection(Line L, Circle C, Point* sol)

{

    Vector nor = normal(L.v);

    Line pl = Line(C.c, nor);

    Point ip = GetIntersection(pl, L);

    double dis = Length(ip - C.c);

    if (dcmp(dis - C.r) > 0) return 0;

    Point dxy = vecunit(L.v) * sqrt(sqr(C.r) - sqr(dis));

    int ret = 0;

    sol[ret] = ip + dxy;

    if (OnSegment(sol[ret], L.p, L.point(1))) ret++;

    sol[ret] = ip - dxy;

    if (OnSegment(sol[ret], L.p, L.point(1))) ret++;

    return ret;

}

int T, n;

vector<Point> sol;

Circle C[30];

Point s, t;

bool vis[maxn];

bool check(Point A, Point B)

{

    vector<Point> gank;

    gank.PB(A);

    gank.PB(B);

    Point roshan[2];

    REP(i, n)

    {

        //线段ab与所有圆的交点

        int m = getSegCircleIntersection(Line(A, B-A), C[i], roshan);

        if(m == 1) gank.PB(roshan[0]);

        if(m == 2) gank.PB(roshan[0]), gank.PB(roshan[1]);

    }

    sort(gank.begin(), gank.end());

    int nc = gank.size();

    REP(i, nc-1)

    {

        Point mid = (gank[i] + gank[i+1]) / 2.0;

        //重点跳过

        if(mid == gank[i]) continue;

        bool flag = 0;

        REP(j, n) if(InCircle(mid, C[j]))//分段中点是否被任意圆覆盖

        {

            flag = 1;

            break;

        }

        if(!flag) return false;

    }

    return true;

}

struct Heap

{

    double d; int u;

    bool operator < (const Heap& rhs) const

    {

        return dcmp(d-rhs.d) > 0;

    }

};

struct Edge

{

    int from, to;

    double dist;

};

vector<Edge> edges;

vector<int> G[maxn];

bool done[maxn];

double d[maxn];

double dij(int s, int t, int n)

{

    priority_queue<Heap> q; q.push((Heap){0.0, s});

    REP(i, n) d[i] = INF; d[s] = 0.0;

    CLR(done, 0);

    while(!q.empty())

    {

        Heap x = q.top(); q.pop();

        int u = x.u, nc = G[u].size();

        if(done[u]) continue;

        done[u] = 1;

        REP(i, nc)

        {

            Edge& e = edges[G[u][i]];

            if(dcmp(d[e.to]-d[u]-e.dist) > 0)

            {

                d[e.to] = d[u] + e.dist;

                q.push((Heap){d[e.to], e.to});

            }

        }

    }

    return d[t];

}

void add(int from, int to, double dist)

{

    edges.PB((Edge){from, to, dist});

    edges.PB((Edge){to, from, dist});

    int nc = edges.size();

    G[from].PB(nc-2);

    G[to].PB(nc-1);

}

void init()

{

    REP(i, maxn) G[i].clear();

    sol.clear();

    edges.clear();

}

int main()

{

    scanf("%d", &T);

    FF(kase, 1, T+1)

    {

        init();

        scanf("%d", &n);

        //sol存平面上所有点

        REP(i, n)

        {

            scanf("%lf%lf%lf", &C[i].c.x, &C[i].c.y, &C[i].r);

            if(i == 0 || i == n-1)

                sol.PB(C[i].c);//加入圆心....

        }

        REP(i, n) FF(j, i+1, n)

            getCCintersection(C[i], C[j], sol);//所有圆的交点

        //去重

        sort(sol.begin(), sol.end());

        int nc = unique(sol.begin(), sol.end()) - sol.begin();

        int s, t;

        REP(i, nc)

        {

            FF(j, i+1, nc) if(check(sol[i], sol[j]))

            //判断两点是否连通

                add(i, j, Length(sol[i] - sol[j]));

            //最短路的起始点

            if(sol[i] == C[0].c) s = i;

            if(sol[i] == C[n-1].c) t = i;

        }

        double ans = dij(s, t, nc);

        printf("Case %d: ", kase);

        if(dcmp(ans-INF) >= 0) puts("No such path.");

        else printf("%.4lf\n", ans);

    }

    return 0;

}