The Doors

The Doors

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. 

Input

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.

Output

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.

Sample Input

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1

Sample Output

10.00
10.06

题目就是说,给你一个10*10的正方形房间,里面会用一些垂直与x轴的墙隔开.每个墙上有两个门,然后给你门的两个端点坐标,要求求出从(0,5)走到(10,5)的最短路；

那其实本质上就是最短路,如果没有墙隔开,就单单这几个点,那就很水了.现在还要求判断一下,某两点之间是否可以直达.所谓直达,就是,这两点的连线不与其他任何墙相交(交于端点是允许的).

至此,题目解完了.

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<cmath>
 #define INF 10000000
 using namespace std;
 ;
 int n;
 ][];
 ];
 ],o;
 double dis(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
 :x>eps;}
 double cross(point a,point b){return a.x*b.y-a.y*b.x;}
 point operator - (point a,point b){point ret; ret.x=a.x-b.x,ret.y=a.y-b.y; return ret;}
 int Intersect(point a,point b,point c,point d){
     double s1,s2,s3,s4;
     int d1,d2,d3,d4;
     d1=dcmp(s1=cross(a-c,b-c));
     d2=dcmp(s2=cross(a-d,b-d));
     d3=dcmp(s3=cross(c-a,d-a));
     d4=dcmp(s4=cross(c-b,d-b));
     &&(d3^d4)==-) ;
     ;
 }
 int main(){
     ; scanf("%d",&n)){
         ,cnt0=; P[].x=,P[].y=;
         ; i<n; i++){
             double x,y1,y2,y3,y4;
             scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
             P[cnt].x=x,P[cnt++].y=y1;
             P[cnt].x=x,P[cnt++].y=y2;
             P[cnt].x=x,P[cnt++].y=y3;
             P[cnt].x=x,P[cnt++].y=y4;
             seg[cnt0].a.x=x,seg[cnt0].a.y=,seg[cnt0].b.x=x,seg[cnt0++].b.y=y1;
             seg[cnt0].a.x=x,seg[cnt0].a.y=y2,seg[cnt0].b.x=x,seg[cnt0++].b.y=y3;
             seg[cnt0].a.x=x,seg[cnt0].a.y=y4,seg[cnt0].b.x=x;seg[cnt0++].b.y=;
         }
         P[cnt].x=,P[cnt].y=;
         ; i<=cnt; i++){
             ; j<=cnt; j++) cost[i][j]=INF; cost[i][i]=;
         }
         ; i<=cnt; i++){
             ; j<=cnt; j++) if (j!=i){
                 o.a.x=P[i].x,o.a.y=P[i].y,o.b.x=P[j].x,o.b.y=P[j].y;
                 ;
                 ; k<cnt0; k++) ){flag=; break;}
                 ) cost[i][j]=dis(P[i],P[j]);
             }
         }
         ; k<=cnt; k++)
             ; i<=cnt; i++)
                 ; j<=cnt; j++) if (cost[i][k]+cost[k][j]<cost[i][j]) cost[i][j]=cost[i][k]+cost[k][j];
         printf(][cnt]);
     }
     ;
 }