HDU 4063 Aircraft --几何，最短路

题意： 给一些圆，要求从第一个圆的圆心走到最后一个圆的圆心，中间路径必须在某个圆内，求最短路径的长度。

解法： 易知要保持在圆内且路径最短，走两圆相交的点能使路径尽量短，所以我们找出所有的两圆相交的点，再加上起点和终点，放到一个容器中，去重后，判断每两点之间的线段是否都在圆内，如果是则建边，建完所有的边后跑一个SPFA即可得出最短路。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

#define Mod 1000000007

#define eps 1e-8

using namespace std;

#define N 100017



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); }

    void input() { scanf("%lf%lf%lf",&c.x,&c.y,&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; }

};

int dcmp(double x) {

    if(x < -eps) return -1;

    if(x > eps) return 1;

    return 0;

}

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; }

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 Angle(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 VectorUnit(Vector x){ return x / Length(x);}

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

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



bool OnSegment(Point P, Point A, Point B) {

    return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) < 0;

}

bool InCircle(Point x, Circle c) { return dcmp(c.r - Length(c.c-x)) > 0; } //not in border

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);

}

Point GetLineIntersection(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 GetCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol) //return 交点个数

{

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

    if(dcmp(d) == 0){

        if(dcmp(C1.r - C2.r) == 0) return -1;  //两圆重合

        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 = angle(C2.c - C1.c);             //向量C1C2的极角

    double da = acos((sqr(C1.r) + sqr(d) - sqr(C2.r)) / (2*C1.r*d)); //C1C2到C1P1的极角



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

    sol.push_back(p1);

    if(p1 == p2) return 1;

    sol.push_back(p2);

    return 2;

}

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

{

    Vector Noml = Normal(L.v);

    Line PL = Line(C.c, Noml);

    Point IP = GetLineIntersection(PL, L); //弦的中点

    double Dis = Length(IP - C.c);

    if(dcmp(Dis-C.r) > 0) return 0;        //在圆外

    Vector HalfChord = VectorUnit(L.v)*sqrt(sqr(C.r)-sqr(Dis));

    int ind = 0;

    sol[ind] = IP + HalfChord;

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

    sol[ind] = IP - HalfChord;

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

    return ind;

}



//data segment

vector<Point> sol;

Circle C[33];

double dis[2510];

vector<pair<int,double> > G[2510];

int n,vis[2510],S,E;

//data ends



bool CheckSegInCircle(Point A, Point B){

    int i,j;

    vector<Point> now;

    now.push_back(A), now.push_back(B);

    Point inter[2];

    for(i=1;i<=n;i++) {

        int m = GetSegCircleIntersection(Line(A,B-A),C[i],inter);

        for(j=1;j<=m;j++) now.push_back(inter[j-1]);

    }

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

    int sz = now.size();

    for(i=0;i<sz-1;i++) {

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

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

        for(j=1;j<=n;j++)

            if(InCircle(mid,C[j]))

                break;

        if(j == n+1) return false;

    }

    return true;

}



void SPFA(int n)

{

    for(int i=0;i<=n;i++) dis[i] = Mod;

    memset(vis,0,sizeof(vis));

    dis[S] = 0, vis[S] = 1;

    queue<int> q;

    q.push(S);

    while(!q.empty())

    {

        int u = q.front();

        q.pop(); vis[u] = 0;

        for(int i=0;i<G[u].size();i++)

        {

            int v = G[u][i].first;

            double w = G[u][i].second;

            if(dis[v] > dis[u] + w)

            {

                dis[v] = dis[u] + w;

                if(!vis[v]) vis[v] = 1, q.push(v);

            }

        }

    }

}



int main()

{

    int t,cs = 1,i,j;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        sol.clear();

        for(i=0;i<=2500;i++) G[i].clear();

        C[1].input(), sol.push_back(C[1].c);

        for(i=2;i<n;i++) C[i].input();

        C[n].input(), sol.push_back(C[n].c);

        for(i=1;i<=n;i++)

            for(j=i+1;j<=n;j++)

                GetCircleCircleIntersection(C[i],C[j],sol);

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

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

        for(i=0;i<ind;i++)

        {

            for(j=i+1;j<ind;j++)

            {

                if(CheckSegInCircle(sol[i],sol[j]))

                {

                    G[i].push_back(make_pair(j,Length(sol[i]-sol[j])));

                    G[j].push_back(make_pair(i,Length(sol[i]-sol[j])));

                }

            }

            if(sol[i] == C[1].c) S = i;

            if(sol[i] == C[n].c) E = i;

        }

        SPFA(ind);

        printf("Case %d: ",cs++);

        if(dcmp(dis[E]-Mod) >= 0) puts("No such path.");

        else printf("%.4f\n",dis[E]);

    }

    return 0;

}

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