1 // 最短路+线段交 POJ 1556 好题  2 // 题意：从(0,5)到(10,5)的最短距离，中间有n堵墙，每堵上有两扇门可以通过  3 // 思路：先存图。直接n^2来暴力，不好写。分成三部分，起点 终点和之间的点：中间点之间：起点和终点的距离  4 // n最大为18所以直接n^3最短路  5   6   7 #include <cstdio>  8 #include <cstring>  9 #include <iostream> 10 #include <algorithm> 11 #include <map> 12 #include <set> 13 #include <queue> 14 #include <stdlib.h> 15 #include <cmath> 16 using namespace std; 17 typedef long long LL; 18 const double inf = 1e9; 19 const int N = 5000; 20 const double eps = 1e-8; 21 void fre() {freopen("in.txt","r",stdin);} 22  23 int sgn(double x){ 24     if(fabs(x)<eps) return  0; 25     if(x<0) return -1; 26     return 1; 27 } 28  29 struct Point{ 30     double x,y; 31     Point(){} 32     Point(double _x,double _y){ 33         x=_x;y=_y; 34     } 35     Point operator -(const Point &b)const{ 36         return Point(x-b.x,y-b.y); 37     } 38     double operator *(const Point &b)const{ 39         return x*b.x+y*b.y; 40     } 41     double operator ^(const Point &b)const{ 42         return x*b.y-y*b.x; 43     }  44 }; 45  46 struct Line{ 47     Point s,e; 48     Line(){} 49     Line(Point _s,Point _e){ 50         s=_s,e=_e; 51     } 52 }; 53 Line line[N]; 54 double xmult(Point p0,Point p1,Point p2){ 55     return (p1-p0)^(p2-p0); 56 } 57  58 double dist(Point a,Point b){ 59     return sqrt((a-b)*(a-b)); 60 } 61  62 bool inter(Line l1,Line l2){ 63     return max(l1.s.x,l1.e.x)>=min(l2.s.x,l2.e.x)&&max(l2.s.x,l2.e.x)>=min(l1.s.x,l1.e.x)&&max(l1.s.y,l1.e.y)>=min(l2.s.y,l2.e.y)&&max(l2.s.y,l2.e.y)>=min(l1.s.y,l1.e.y)&&sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e))<=0&&sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e))<=0; 64 } 65  66 double d[100][100]; 67 int main(){ 68     // fre(); 69     int n; 70     while(~scanf("%d",&n)){ 71         if(n==-1) break; 72         // clc(d, 73         for(int i=0;i<=4*n+1;i++){ 74             for(int j=0;j<=4*n+1;j++){ 75                 if(i==j) d[i][j]=0; 76                 else d[i][j]=inf; 77             } 78         } 79         for(int i=1;i<=n;i++){ 80             double x,y3,y1,y4,y2; 81             scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4); 82             line[i*2-1]=Line(Point(x,y1),Point(x,y2)); 83             line[i*2]=Line(Point(x,y3),Point(x,y4)); 84         } 85         bool flag; 86         for(int i=1;i<=4*n;i++){ 87             flag=true; 88             int cn=(i+3)/4; 89             Point tem; 90             if(i&1) tem=line[(i+1)/2].s; 91             else tem=line[(i+1)/2].e; 92             for(int j=1;j<cn;j++){ 93                 if(inter(line[j*2-1],Line(Point(0,5),tem))==false&&inter(line[j*2],Line(Point(0,5),tem))==false){ 94                     flag=false; 95                     break; 96                 } 97             } 98             if(flag) d[0][i]=d[i][0]=dist(Point(0,5),tem); 99             flag=true;100             for(int j=cn+1;j<=n;j++){101                 if(inter(line[j*2-1],Line(Point(10,5),tem))==false&&inter(line[j*2],Line(Point(10,5),tem))==false){102                     flag=false;103                     break;104                 }105             }106             if(flag) d[4*n+1][i]=d[i][4*n+1]=dist(Point(10,5),tem);107         }108         flag=true;109         for(int i=1;i<=4*n;i++){110             for(int j=i+1;j<=4*n;j++){111                flag=true;112                Point p1,p2;113                int cn1,cn2;114                cn1=(i+3)/4;115                cn2=(j+3)/4;116                if(i&1) p1=line[(i+1)/2].s;117                else p1=line[(i+1)/2].e;118                if(j&1) p2=line[(j+1)/2].s;119                else p2=line[(j+1)/2].e;120                for(int cn=cn1+1;cn<cn2;cn++){121                     if(inter(line[cn*2-1],Line(p1,p2))==false&&inter(line[cn*2],Line(p1,p2))==false){122                       flag=false;123                       break;124                     }125                }126                if(flag) d[i][j]=d[j][i]=dist(p1,p2);127             }128         }129         flag=true;130         for(int i=1;i<=n;i++){131             if(inter(line[i*2-1],Line(Point(0,5),Point(10,5)))==false&&inter(line[i*2],Line(Point(0,5),Point(10,5)))==false){132                 flag=false;133                 break;134             }135         }136         if(flag) d[0][4*n+1]=d[4*n+1][0]=dist(Point(0,5),Point(10,5));137         for(int i=0;i<=4*n+1;i++){138             for(int j=0;j<=4*n+1;j++){139                 for(int k=0;k<=4*n+1;k++){140                     if(d[i][k]+d[k][j]<d[i][j]) d[i][j]=d[i][k]+d[k][j];141                 }142             }143         }144         printf("%.2f\n",d[0][4*n+1]);145     }146     return 0;147 }