You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5). There will also be from 0 to 18 vertical walls inside the chamber, each with two doorways. The figure below illustrates such a chamber and also shows the path of minimal length. 

The input data for the illustrated chamber would appear as follows.

2 
4 2 7 8 9 
7 3 4.5 6 7

The first line contains the number of interior walls. Then there is a line for each such wall, containing five real numbers. The first number is the x coordinate of the wall (0 < x < 10), and the remaining four are the y coordinates of the ends of the doorways in that wall. The x coordinates of the walls are in increasing order, and within each line the y coordinates are in increasing order. The input file will contain at least one such set of data. The end of the data comes when the number of walls is -1.

The output should contain one line of output for each chamber. The line should contain the minimal path length rounded to two decimal places past the decimal point, and always showing the two decimal places past the decimal point. The line should contain no blanks.

1

5 4 6 7 8

2

4 2 7 8 9

7 3 4.5 6 7

-1

10.00

10.06

#include<cstdio>

#include<cstring>

#include<cmath>

#include<vector>

#include<algorithm>

#include<queue>

#define MAXN 4*20

#define INF 0x3f3f3f3f

using namespace std;

//--------------------------------------计算几何模板 - st--------------------------------------

const double eps = 1e-;

struct Point{

    double x,y;

    Point(double tx=,double ty=):x(tx),y(ty){}

};

typedef Point Vctor;

//向量的加减乘除

Vctor operator + (Vctor A,Vctor B){return Vctor(A.x+B.x,A.y+B.y);}

Vctor operator - (Point A,Point B){return Vctor(A.x-B.x,A.y-B.y);}

Vctor operator * (Vctor A,double p){return Vctor(A.x*p,A.y*p);}

Vctor operator / (Vctor A,double p){return Vctor(A.x/p,A.y/p);}

int dcmp(double x)

{

    if(fabs(x)<eps) return ;

    else return (x<)?(-):();

}

bool operator == (Point A,Point B){return dcmp(A.x-B.x)== && dcmp(A.y-B.y)==;}

//向量的点积，长度，夹角

double Dot(Vctor A,Vctor B){return A.x*B.x+A.y*B.y;}

double Length(Vctor A){return sqrt(Dot(A,A));}

double Angle(Vctor A,Vctor B){return acos(Dot(A,B)/Length(A)/Length(B));}

//叉积，三角形面积

double Cross(Vctor A,Vctor B){return A.x*B.y-A.y*B.x;}

double TriangleArea(Point A,Point B,Point C){return Cross(B-A,C-A);}

//判断线段是否规范相交

bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)

{

    double c1 = Cross(a2 - a1,b1 - a1), c2 = Cross(a2 - a1,b2 - a1),

           c3 = Cross(b2 - b1,a1 - b1), c4 = Cross(b2 - b1,a2 - b1);

    return dcmp(c1)*dcmp(c2)< && dcmp(c3)*dcmp(c4)<;

}

//--------------------------------------计算几何模板 - ed--------------------------------------

//--------------------------------------spfa算法 - st--------------------------------------

double d[MAXN];

double mp[MAXN][MAXN];

bool vis[MAXN];

void init(int n){for(int i=;i<n;i++) for(int j=;j<n;j++) mp[i][j]=;}

void addedge(int u,int v,double w){mp[u][v]=w;}

void spfa(int st,int n)

{

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

    {

        i==st ? d[i]= : d[i]=INF;

        vis[i]=;

    }

    queue<int> q;

    q.push(st);

    vis[st]=;

    while(!q.empty())

    {

        int u=q.front();q.pop();vis[u]=;

        for(int v=;v<n;v++)

        {

            if(u==v || mp[u][v]==) continue;

            double tmp=d[v];

            if(d[v]>d[u]+mp[u][v]) d[v]=d[u]+mp[u][v];

            if(d[v]<tmp && !vis[v])

            {

                q.push(v);

                vis[v]=;

            }

        }

    }

}

//--------------------------------------spfa算法 - ed--------------------------------------

int n;

vector<Point> p;

int main()

{

    while(scanf("%d",&n) && n!=-)

    {

        p.clear();

        p.push_back(Point(,));

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

        {

            double x,y1,y2,y3,y4;

            scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);

            p.push_back(Point(x,y1));

            p.push_back(Point(x,y2));

            p.push_back(Point(x,y3));

            p.push_back(Point(x,y4));

        }

        p.push_back(Point(,));

        int _size=p.size();

        init(_size);

        for(int i=;i<_size;i++)

        {

            for(int j=i+;j<_size;j++)

            {

                if(p[i].x==p[j].x) continue;

                bool ok=;

                for(int k=i+;k<j;k++)

                {

                    if(k%==)

                    {

                        if(SegmentProperIntersection(p[i],p[j],p[k],Point(p[k].x,)))

                        {

                            ok=;

                            break;

                        }

                    }

                    else if(k%==)

                    {

                        if(SegmentProperIntersection(p[i],p[j],p[k],p[k+]))

                        {

                            ok=;

                            break;

                        }

                    }

                    else if(k%==)

                    {

                        if(SegmentProperIntersection(p[i],p[j],p[k],p[k-]))

                        {

                            ok=;

                            break;

                        }

                    }

                    else if(k%==)

                    {

                        if(SegmentProperIntersection(p[i],p[j],p[k],Point(p[k].x,)))

                        {

                            ok=;

                            break;

                        }

                    }

                }

                if(ok) addedge(i,j,Length(p[i]-p[j]));

            }

        }

        spfa(,_size);

        printf("%.2f\n",d[_size-]);

    }

}