hdu 4063 Aircraft(计算几何+最短路)

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;
}