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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-8using namespace std;#define N 100017struct 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 borderdouble 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 segmentvector<Point> sol;Circle C[33];double dis[2510];vector<pair<int,double> > G[2510];int n,vis[2510],S,E;//data endsbool 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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HDU 4063 Aircraft --几何,最短路