poj 1556 The Doors 线段相交，最短路

来源：互联网发布：pc录屏软件编辑：程序博客网时间：2024/05/16 00:52

题目地址：poj1556

首先，这是个最短路问题，数据范围比较小，可以考虑floyd

首先不难想象，最后的路径，只要是拐点，那一定是经过某个门的端点的，否则稍微扰动就可以变大变小。

对于两点间路径的长度，他们连结线段被某条线段中断，那么他们的距离应该是通过连接某几个中间点完成的。否则，没有中断，那就是欧氏距离。

初始化以后，然后floyd一遍就可以了。

代码：

#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>const double eps=1e-8;using namespace std;struct Point{    double x;    double y;    Point(double x=0,double y=0):x(x),y(y){}    void operator<<(Point &A) {cout<<A.x<<' '<<A.y<<endl;}};int dcmp(double x)  {return (x>eps)-(x<-eps); }typedef  Point  Vector;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);}// ps  coutostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y<<endl; return out;}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 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  Cross(Vector A,Vector B)  {return A.x*B.y-B.x*A.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  Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);}Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){    Vector u=P-Q;    double t=Cross(w, u)/Cross(v,w);    return P+v*t;    }double DistanceToLine(Point P,Point A,Point B){    Vector v1=P-A; Vector v2=B-A;    return fabs(Cross(v1,v2))/Length(v2);    }double DistanceToSegment(Point P,Point A,Point B){    if(A==B)  return Length(P-A);        Vector v1=B-A;    Vector v2=P-A;    Vector v3=P-B;        if(dcmp(Dot(v1,v2))==-1)    return  Length(v2);    else if(Dot(v1,v3)>0)    return Length(v3);        else return DistanceToLine(P, A, B);    }Point GetLineProjection(Point P,Point A,Point B){    Vector v=B-A;    Vector v1=P-A;    double t=Dot(v,v1)/Dot(v,v);        return  A+v*t;}bool  SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){    double c1=Cross(b1-a1, a2-a1);    double c2=Cross(b2-a1, a2-a1);    double c3=Cross(a1-b1, b2-b1);    double c4=Cross(a2-b1, b2-b1);            return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ;    }bool  OnSegment(Point P,Point A,Point B){    return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;}double PolygonArea(Point *p,int n){    double area=0;        for(int i=1;i<n-1;i++)    {        area+=Cross(p[i]-p[0], p[i+1]-p[0]);            }    return area/2;    }Point  read_point(){    Point P;    scanf("%lf%lf",&P.x,&P.y);    return  P;}const double INF=0x3fffffff;double d[100][100];Point   p[100];int  n;int N;void  init(){    for(int i=0;i<100;i++)        for(int j=0;j<100;j++)            d[i][j]=(i==j)?0:INF;        }void floyd(){  for(int k=0;k<N;k++)     for(int i=0;i<N;i++)        for(int j=0;j<N;j++)        {            d[i][j]=min(d[i][j],d[i][k]+d[k][j]);        }}int main(){    double h[5];          while(cin>>n)    {        if(n==-1) break;                init();               for(int i=0;i<n;i++)        {            for(int j=0;j<5;j++)              scanf("%lf",&h[j]);                        for(int j=0;j<4;j++)             p[4*i+j]=Point(h[0],h[j+1]);                }                N=4*n;;                p[N++]=Point(0,5);        p[N++]=Point(10,5);        //        for(int i=0;i<N;i++)//            cout<<p[i];                for(int i=0;i<N;i++)            for(int j=0;j<N;j++)            {                bool ok=1;                                for(int k=0;k<n;k++)                {                    Point bottom=Point(p[4*k].x,0);                    Point top=Point(p[4*k].x,10);                                        if(SegmentProperIntersection(p[i], p[j], bottom, p[4*k])){                                               ok=0;                    }                                        if(SegmentProperIntersection(p[i], p[j], p[4*k+1], p[4*k+2])){                                             ok=0;                    }                                        if(SegmentProperIntersection(p[i], p[j], p[4*k+3], top)){                        ok=0;                    }                                                                            }                                              if(ok) d[i][j]=Length(p[i]-p[j]);            }                        floyd();                 printf("%.2f\n",d[N-2][N-1]);            }}

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