hdu 4063 福州赛区网络赛 圆 ****

画几个图后，知道路径点集一定是起点终点加上圆与圆之间的交点，枚举每两个点之间是否能走，能走则连上线，然后求一遍最短路即可

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define sqr(x) ((x)*(x))
const double eps= 1e-6;
const int maxc=30;
const int maxp=1000;
const double inf=1e20;
double dsgn(double x)//return double
{
    if(fabs(x)<eps)return 0;
    return x;
}
int isgn(double x)//return int
{
    if(fabs(x)<eps)return 0;
    if(x<0)return -1;
    return 1;
}
struct point
{
    double x, y;
    point() {}
    point(double xx, double yy):x(xx), y(yy) {}
    double length() const
    {
        return sqrt(sqr(x)+sqr(y));
    }
    point set(const double &m) const
    {
        double len=length();
        return point(x*m/len, y*m/len);
    }
    double sqrdist(const point &a)const
    {
        return sqr(a.x-x)+sqr(a.y-y);
    }
    double dist(const point &a)const
    {
        return sqrt(sqrdist(a));
    }
} p[maxp];
struct line
{
    double a, b, c;
    line() {}
    line(double ta=1,  double tb=-1,  double tc=0):a(ta), b(tb), c(tc) {}
};
struct circle
{
    point o;
    double r;
} c[maxc];
int pn, cn;
double edge[maxp][maxp];

int cir_relation(circle c1, circle c2)//判断两圆关系
{
    double d=c1.o.dist(c2.o);
    int a=isgn(d-c1.r-c2.r);
    int b=isgn(d-fabs(c1.r-c2.r));
    if(a==0)return -2;//外切
    if(b==0)return -1;//内切
    if(a>0)return 2;//相离
    if(b<0)return 1;//内含
    return 0;//相交
}
point line_intersect(point p1, point p2, point q1, point q2)
{
    point ans=p1;
    double t=((p1.x-q1.x)*(q1.y-q2.y)-(p1.y-q1.y)*(q1.x-q2.x))/
             ((p1.x-p2.x)*(q1.y-q2.y)-(p1.y-p2.y)*(q1.x-q2.x));
    ans.x += (p2.x-p1.x)*t;
    ans.y += (p2.y-p1.y)*t;
    return ans;
}
void line_cross_circle(point p, point q, point o, double r, point &pp, point &qq)
{
    point tmp=o;
    tmp.x += p.y-q.y;
    tmp.y += q.x-p.x;
    tmp=line_intersect(tmp, o, p, q);
    double t=sqrt(r*r - sqr(tmp.dist(o)))/(p.dist(q));
    pp.x=tmp.x + (q.x-p.x)*t;
    pp.y=tmp.y + (q.y-p.y)*t;
    qq.x=tmp.x - (q.x-p.x)*t;
    qq.y=tmp.y - (q.y-p.y)*t;
}
void circle_intersect(circle c1, circle c2, point &p1, point &p2)
{
    point u, v;
    double t, d;
    d=c1.o.dist(c2.o);
    t= (1+ (sqr(c1.r) -sqr(c2.r))/sqr(d))/2;
    u.x= c1.o.x+ (c2.o.x-c1.o.x)*t;
    u.y= c1.o.y+ (c2.o.y-c1.o.y)*t;
    v.x= u.x+c1.o.y-c2.o.y;
    v.y= u.y-c1.o.x+c2.o.x;
    line_cross_circle(u, v, c1.o, c1.r, p1, p2);
}
point circle_tangent(circle c1, circle c2)
{
    point t;
    if(isgn(c1.o.dist(c2.o)-c1.r-c2.r)==0)
    {
        t.x=(c1.r*c2.o.x + c2.r*c1.o.x)/(c1.r+c2.r);
        t.y=(c1.r*c2.o.y + c2.r*c1.o.y)/(c1.r+c2.r);
        return t;
    }
    t.x=(c1.r*c2.o.x-c2.r*c1.o.x)/(c1.r-c2.r);
    t.y=(c1.r*c2.o.y-c2.r*c1.o.y)/(c1.r-c2.r);
    return t;
}
void findallpoint()
{
    int i, j, rel;
    point p1, p2;
    for(i=1; i<=cn; i++)
        for(j=i+1; j<=cn; j++)
        {
            rel=cir_relation(c[i], c[j]);
            if(rel==0)
            {
                circle_intersect(c[i], c[j], p1, p2);
                p[pn++]=p1;
                p[pn++]=p2;
            }
            else if(rel<0)
                p[pn++]=circle_tangent(c[i], c[j]);
        }
}

point tmp[100];
point qq[3], base;
bool cmp(point a, point b)
{
    return a.dist(base)<b.dist(base);
}
bool insideok(point a, point b)
{
    for(int i=1; i<=cn; i++)
    {
        if(isgn(c[i].r-a.dist(c[i].o))<0)continue;
        if(isgn(c[i].r-b.dist(c[i].o))<0)continue;
        return true;
    }
    return false;
}
double multiply(point sp, point ep, point op)
{
    return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x);
}
bool onsegment(point a, point u, point v)
{
    return isgn(multiply(v, a, u))==0 && isgn((u.x-a.x)*(v.x-a.x))<=0 && isgn((u.y-a.y)*(v.y-a.y))<=0;
}

bool edgeok(point a, point b)
{
    int i, j;
    int cnt=0, num;

    tmp[cnt++]=a;
    point p1, p2;
    for(i=1; i<=cn; i++)
    {
        line_cross_circle(a, b, c[i].o, c[i].r, p1, p2);
        if(onsegment(p1, a, b))tmp[cnt++]=p1;
        if(onsegment(p2, a, b))tmp[cnt++]=p2;
    }
    tmp[cnt++]=b;

    base=a;
    sort(tmp, tmp+cnt, cmp);
    for(i=1; i<cnt; i++)
        if(!insideok(tmp[i-1], tmp[i]))
            return false;
    return true;
}
void findalledge()
{
    int i, j;
    for(i=0; i<pn; i++)
        for(j=i+1; j<pn; j++)
        {
            if(edgeok(p[i], p[j]))
                edge[i][j]=edge[j][i]=p[i].dist(p[j]);
            else
                edge[i][j]=edge[j][i]=-1;
        }
}
bool has[maxp];
double dis[maxp];
void dijkstra()
{
    int i, j, num;
    double tmp;
    for(i=0; i<pn; i++)has[i]=false, dis[i]=inf;
    dis[0]=0;
    for(i=0; i<pn; i++)
    {
        tmp=inf;
        for(j=0; j<pn; j++)
            if(!has[j] && dis[j]<tmp)
            {
                tmp=dis[j];
                num=j;
            }
        has[num]=true;
        for(j=0; j<pn; j++)
            if(!has[j] && isgn(edge[num][j])>=0)
                dis[j]=min(dis[j], dis[num]+edge[num][j]);
    }
    if(dis[1]<inf)printf("%.4lf
", dis[1]);
    else printf("No such path.
");
}

void solve()
{
    findallpoint();
    findalledge();
    dijkstra();
}

int main()
{
    int t, ca=1;
    scanf("%d", &t);
    while(t--0)
    {
        scanf("%d", &cn);
        for(int i=1; i<=cn; i++)
            scanf("%lf%lf%lf", &c[i].o.x, &c[i].o.y, &c[i].r);
        pn=0;
        p[pn++]=c[1].o;
        p[pn++]=c[cn].o;
        printf("Case %d: ", ca++);
        solve();
    }
    return 0;
}