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;
}