描述

坐标系,x,y轴都是0~10.起点(0,5),终点(10,5),中间可能有墙,每一堵墙有两个门,给出门的上下定点的坐标,求从起点到终点的最短路.

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 file 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

分析

注意:

1.不能用小写的vector(如果有vector的头文件的话).

2.在segment_cross_simple函数中括号要特!别!小!心!调了一个多小时...

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 const int maxn=100;
 5 const double oo=~0u>>1,eps=1e-10;
 6 int wall_num,edge_num,point_num;
 7 double d[maxn][maxn];
 8
 9 struct Point{
10     double x,y;
11     Point(double x=0,double y=0):x(x),y(y){}
12 }p[maxn];
13 typedef Point Vector;
14 Vector operator + (Vector a,Vector b){ return Vector(a.x+b.x,a.y+b.y); }
15 Vector operator - (Vector a,Vector b){ return Vector(a.x-b.x,a.y-b.y); }
16 Vector operator * (Vector a,double p){ return Vector(a.x*p,a.y*p); }
17 Vector operator / (Vector a,double p){ return Vector(a.x/p,a.y/p); }
18 struct edge{
19     double x1,y1,x2,y2;
20     edge(double x1=0,double y1=0,double x2=0,double y2=0):x1(x1),y1(y1),x2(x2),y2(y2){}
21 }g[maxn];
22 void add_point(double x,double y){
23     p[++point_num]=Point(x,y);
24 }
25 void add_edge(double x1,double y1,double x2,double y2){
26     g[++edge_num]=edge(x1,y1,x2,y2);
27 }
28 inline int dcmp(double x){
29     if(fabs(x)<eps) return 0;
30     return x<0?-1:1;
31 }
32 inline double cross(Vector a,Vector b){
33     return a.x*b.y-a.y*b.x;
34 }
35 inline bool segment_cross_simple(Point a,Point b,Point c,Point d){
36 return (dcmp(cross(b-a,c-a))^dcmp(cross(b-a,d-a)))==-2&&(dcmp(cross(d-c,a-c))^dcmp(cross(d-c,b-c)))==-2;
37 }
38 inline double dis(Point a,Point b){
39     return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));
40 }
41 void Floyd(){
42     for(int k=1;k<=point_num;k++)
43         for(int i=1;i<=point_num;i++)
44             for(int j=1;j<=point_num;j++)
45                 d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
46 }
47 void outit(){
48     for(int i=1;i<=point_num;i++){
49         for(int j=1;j<=point_num;j++){
50             printf("d[%d][%d]=%.2lf\t",i,j,d[i][j]);
51         }
52         printf("\n");
53     }
54 }
55
56 void solve(){
57     for(int i=1;i<=point_num;i++)
58         for(int j=i+1;j<=point_num;j++){
59             bool link=true;
60             for(int k=1;k<=edge_num;k++){
61                 Point t1=Point(g[k].x1,g[k].y1);
62                 Point t2=Point(g[k].x2,g[k].y2);
63                 if(segment_cross_simple(p[i],p[j],t1,t2)){
64                     link=false;
65                     break;
66                 }
67             }
68             if(link) d[i][j]=d[j][i]=dis(p[i],p[j]);
69         }
70     Floyd();
71     printf("%.2lf\n",d[1][2]);
72 }
73 int main(){
74     while(scanf("%d",&wall_num)&&wall_num!=-1){
75         point_num=edge_num=0;
76         add_point(0,5);
77         add_point(10,5);
78         for(int i=1;i<=wall_num;i++){
79             double x,y1,y2,y3,y4;
80             scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
81             add_point(x,y1); add_point(x,y2); add_point(x,y3); add_point(x,y4);
82             add_edge(x,0,x,y1); add_edge(x,y2,x,y3); add_edge(x,y4,x,10);
83         }
84         for(int i=1;i<=point_num;i++){
85             for(int j=1;j<=point_num;j++)
86                 d[i][j]=oo;
87             d[i][i]=0;
88         }
89         solve();
90     }
91     return 0;
92 }