Poj 2502 Subway

                                                                                                  Subway
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 4511   Accepted: 1473

Description

You have just moved from a quiet Waterloo neighbourhood to a big, noisy city. Instead of getting to ride your bike to school every day, you now get to walk and take the subway. Because you don't want to be late for class, you want to know how long it will take you to get to school. 
You walk at a speed of 10 km/h. The subway travels at 40 km/h. Assume that you are lucky, and whenever you arrive at a subway station, a train is there that you can board immediately. You may get on and off the subway any number of times, and you may switch between different subway lines if you wish. All subway lines go in both directions.

Input

Input consists of the x,y coordinates of your home and your school, followed by specifications of several subway lines. Each subway line consists of the non-negative integer x,y coordinates of each stop on the line, in order. You may assume the subway runs in a straight line between adjacent stops, and the coordinates represent an integral number of metres. Each line has at least two stops. The end of each subway line is followed by the dummy coordinate pair -1,-1. In total there are at most 200 subway stops in the city.

Output

Output is the number of minutes it will take you to get to school, rounded to the nearest minute, taking the fastest route.

Sample Input

0 0 10000 1000
0 200 5000 200 7000 200 -1 -1 
2000 600 5000 600 10000 600 -1 -1

Sample Output

21

Source

Waterloo local 2001.09.22
 
 
思路：本题的题意是，给出你家的地点位置（用坐标<X,Y>表示），然后给出你的学校的位置。然后给出一系列地铁站的坐标，一行上的站都是一条线路上的。然后相邻的地铁站之间的距离是相邻的两点之间的直线距离。步行的速度是10KM/H，地铁的速度是40KM/H，问从家到学校的最短时间。
        首先可以求出每两个点之间的通过步行或地铁的最短时间。 然后就是最短路径了。
       注意：这个题目需要注意地铁站之间的距离，使用地铁速度时，只有相邻的地铁站之间距离才是两点之间的直线距离。其他间接点之间的时间，如果要使用地铁的时间都要间接的通过其他的点。否则，使用直接距离就只能使用步行速度了。然后就是存储这些地点信息，地点只有200+2个，但是如果这些点可能会出现在多条地铁线上。所以要不能够将每一条地铁线上的站点都存一次，因为这样很可能会导致许多站点重复存储，甚至会使存储空间超出预期的空间。通过上述方法把距离换算成时间后，然后就用最短路径的解法就可以求出解了。
 
 
 
 
 
 
  1 #include <cstdlib>
  2 #include <iostream>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <cmath>
  6 #include <cctype>
  7 #include <string>
  8 
  9 
 10 #define MAXD 1.0e20
 11 
 12 
 13 using namespace std;
 14 
 15 
 16 
 17 
 18 struct CityStops{
 19        int x,y;
 20        }cityStops[200+5];
 21 
 22 
 23 
 24        
 25        
 26 double d[200+5][200+5];
 27 int vis[200+5];
 28 
 29 double lowcost[200+5];
 30 
 31 
 32 
 33 
 34        
 35 int pos=0;
 36 
 37 int q[205];
 38 int Len=0;
 39 
 40 
 41 int search(int x,int y,int n)
 42 {
 43     int i;
 44     for(i=0;i<n;i++)
 45     {
 46                     if((cityStops[i].x==x)&&(cityStops[i].y==y))
 47                     return i;
 48     }
 49     return -1;
 50 }
 51 
 52 
 53 
 54     
 55 
 56 
 57 int main(int argc, char *argv[])
 58 {
 59     int i,j,k;
 60     int n;
 61     pos=0;
 62     
 63     //int beginflag,endflag;
 64     
 65     
 66     
 67     
 68     
 69     for(i=0;i<=204;i++)
 70     for(j=0;j<=204;j++)
 71     {d[i][j]=MAXD;}
 72     
 73     
 74     
 75     Len=0;
 76     
 77     scanf("%d%d",&cityStops[pos].x,&cityStops[pos].y);
 78     pos++;
 79     scanf("%d%d",&cityStops[pos].x,&cityStops[pos].y);
 80     if(search(cityStops[pos].x,cityStops[pos].y,pos)==-1)
 81     pos++;
 82     
 83     
 84     
 85     while(scanf("%d%d",&cityStops[pos].x,&cityStops[pos].y)!=EOF)
 86     {
 87                                                                if((cityStops[pos].x==-1)&&(cityStops[pos].y==-1))
 88                                                                {break;}
 89                                                                Len=0;
 90                                                                
 91                                                                
 92                                                                //beginflag=endflag=pos;
 93                                                                int tmp=search(cityStops[pos].x,cityStops[pos].y,pos);
 94                                                                if(tmp==-1)
 95                                                                {q[Len++]=pos;pos++;}
 96                                                                else
 97                                                                {q[Len++]=tmp;}
 98                                                                
 99                                                                
100                                                                
101                                                                while(scanf("%d%d",&cityStops[pos].x,&cityStops[pos].y)!=EOF)
102                                                                {
103                                                                if((cityStops[pos].x==-1)&&(cityStops[pos].y==-1))
104                                                                {break;}
105                                                                
106                                                                
107                                                                
108                                                                
109                                                                tmp=search(cityStops[pos].x,cityStops[pos].y,pos);
110                                                                if(tmp==-1)
111                                                                {q[Len++]=pos;pos++;}
112                                                                else
113                                                                {q[Len++]=tmp;}
114                                                                
115                                                                
116                                                                
117                                                                
118                                                                
119                                                                
120                                                                
121                                                                
122                                                                
123                                                                }
124                                                                
125                                                                int ii,jj;
126                                                                for(ii=0;ii<Len-1;ii++)
127                                                                //for(jj=ii+1;jj<Len;jj++)
128                                                                {i=q[ii];
129                                                                 j=i+1;
130                                                                 d[i][j]=d[j][i]=sqrt(1.0*(cityStops[i].x-cityStops[j].x)*(cityStops[i].x-cityStops[j].x)+(cityStops[i].y-cityStops[j].y)*(cityStops[i].y-cityStops[j].y));
131                                                                
132                                                                d[i][j]=d[i][j]*60/40000;
133                                                                d[j][i]=d[i][j];
134                                                                
135                                                                }
136                                                                
137                                                                
138                                                                
139                                                                
140 
141 
142   }
143   
144   
145   
146   
147   n=pos;
148   
149   
150   
151   for(i=0;i<n;i++)
152   for(j=0;j<n;j++)
153   {
154                   if(i==j)
155                   continue;
156                   
157                   if(d[i][j]<MAXD)
158                   continue;
159                   
160                   d[i][j]=sqrt(1.0*(cityStops[i].x-cityStops[j].x)*(cityStops[i].x-cityStops[j].x)+(cityStops[i].y-cityStops[j].y)*(cityStops[i].y-cityStops[j].y));
161                   d[i][j]=d[i][j]*60/10000;
162   }
163   
164   
165 
166 
167 
168 
169 
170 
171 double mincost=MAXD;
172 
173 for(i=0;i<n;i++)
174 {
175                 vis[i]=0;
176                 lowcost[i]=d[0][i];
177 }
178 
179 
180 
181 vis[0]=1;
182 
183 
184 
185 for(i=1;i<n;i++)
186 {
187                 k=-1;
188                 mincost=MAXD;
189                 
190                 for(j=0;j<n;j++)
191                 {
192                                 if((vis[j]==0)&&(mincost>lowcost[j])&&(lowcost[j]<MAXD))
193                                 {k=j;mincost=lowcost[j];}
194                 }
195                 
196                 vis[k]=1;
197                 
198                 for(j=0;j<n;j++)
199                 {
200                                 if((vis[j]==0)&&(lowcost[j]>lowcost[k]+d[k][j]))
201                                 {lowcost[j]=lowcost[k]+d[k][j];}
202                 }
203                 
204 }
205 
206 //for(i=0;i<n;i++)
207 
208 printf("%.0f\n",lowcost[1]);
209 
210                                                                      
211                                 
212                                 
213                                  
214                                                                                                                           
215     
216     
217     
218     
219     
220     
221     
222     
223     
224     
225     
226     //system("PAUSE");
227     return EXIT_SUCCESS;
228 }

 

posted @ 2012-10-03 21:45  cseriscser  阅读(289)  评论(0)  编辑  收藏  举报