POJ 2502 Subway-經過預處理的最短路
阿新 • • 發佈：2017-07-11
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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.
這道題的思路簡單來說就是將所有地鐵站轉換成點，然後預處理所有點間的邊權（即將距離轉換成時間），然後用最短路算法（例如SPFA）求出起點和目標點的距離。
第二： 每點間的距離處理。 你需要判斷他們間的路是在地鐵裏還是人行道，所以我加了一個初始為1的變量p，每次輸入-1，-1是就p++。 然後每次在兩點間賦權的時候判斷是不是在同一條地鐵線上。 第三：
現在還沒解決的問題，如果兩條地鐵線交叉，那麽賦權就又有問題了，所以我希望可以用坐標來表示點，但是由於坐標可能給的很大，所以一直不知道怎麽處理。 附上還沒完成的代碼（甚至樣例都是錯的），希望可以有所啟發（還會不斷更新，知道附上AC代碼）：
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這道題的思路簡單來說就是將所有地鐵站轉換成點，然後預處理所有點間的邊權（即將距離轉換成時間），然後用最短路算法（例如SPFA）求出起點和目標點的距離。
第二： 每點間的距離處理。 你需要判斷他們間的路是在地鐵裏還是人行道，所以我加了一個初始為1的變量p，每次輸入-1，-1是就p++。 然後每次在兩點間賦權的時候判斷是不是在同一條地鐵線上。 第三：
現在還沒解決的問題，如果兩條地鐵線交叉，那麽賦權就又有問題了，所以我希望可以用坐標來表示點，但是由於坐標可能給的很大，所以一直不知道怎麽處理。 附上還沒完成的代碼（甚至樣例都是錯的），希望可以有所啟發（還會不斷更新，知道附上AC代碼）：
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> #define ll long long using namespace std; int read() { int x=0,y=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘) { if(ch==‘-‘) y=-1; ch=getchar(); } while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘; ch=getchar(); } return x*y; } int abs(int x) { if(x<0) return -x; else return x; } int change(double x) { int r=(int)x; if(x-r<0.5) return r; else return r+1; } double way(int x1,int y1,int x2,int y2) { int r1=abs(x1-x2),r2=abs(y1-y2); return sqrt(r1*r1+r2*r2); } struct edge { int next,to; double lon; } e[4045]; int map[305][305],num,node[305][3],head[6045],cnt,t[345],headd,tail=1; double dist[345]; bool vis[345]; void add(int from,int to,double lon) { e[++cnt].lon=lon; e[cnt].to=to; e[cnt].next=head[from]; head[from]=cnt; } int main() { memset(head,-1,sizeof(head)); int x1=read(),y1=read(),x2=read(),y2=read(),x,y,p=1,l=change(way(x1,y1,x2,y2)/1000*6); node[++num][0]=x1; node[num][1]=y1; node[++num][0]=x2; node[num][1]=y2; add(1,2,l); add(2,1,l); while(scanf("%d%d",&x,&y)!=EOF) { if(x==-1&&y==-1) { p++; continue; } node[++num][0]=x; node[num][1]=y; node[num][2]=p; for(int i=1; i<num; i++) { double dis; if(node[i][2]==p) dis=way(x,node[i][0],y,node[i][1])/4000*6; else dis=way(x,node[i][0],y,node[i][1])/1000*6; // printf("x=%d y=%d node[i][0]=%d node[i][1]=%d dis=%f\n",x,y,node[i][0],node[i][1],dis); add(num,i,dis),add(i,num,dis); } } for(int i=2; i<=num; i++) dist[i]=2e8; t[0]=1; while(headd!=tail) { int r=head[t[headd]]; vis[t[headd]]=0; while(r!=-1) { if(dist[e[r].to]>dist[t[headd]]+e[r].lon) { dist[e[r].to]=dist[t[headd]]+e[r].lon; printf("e[r].lon=%f e[r].to=%d dist=%f %f\n",e[r].lon,e[r].to,dist[t[headd]],dist[e[r].to]); if(!vis[e[r].to]) { vis[e[r].to]=1; t[tail++]=e[r].to; } } r=e[r].next; } headd++; } printf("%f",dist[2]); return 0; } // FOR C.H
附上最新經譚姐啟發寫的正解方法代碼（思路稍後，仍有問題在調試）：
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int head[4045],cnt,num; struct edge { int next,to; double lon; } e[4045]; void add(int from,int to,double lon) { e[++cnt].lon=lon; e[cnt].to=to; e[cnt].next=head[from]; head[from]=cnt; } int abs(int x) { if(x<0) return -x; else return x; } double far(int x1,int y1,int x2,int y2) { int r1=abs(x1-x2),r2=abs(y1-y2); return sqrt(r1*r1+r2*r2); } int change(double x) { int r=(int)x; if(x-r<0.5) return r; else return r+1; } int sub[345][2],map[345][2]; void scan() { int x,y,p=0,s=0; while(scanf("%d%d",&x,&y)!=EOF) { if(x==-1&&y==-1) { for(int i=2; i<=s; i++) { double f=far(map[num-s+i-1][0],map[num-s+i-1][1],map[num-s+i][0],map[num-s+i][1])/4000*6; add(num-s+i-1,num-s+i,f); add(num-s+i,num-s+i-1,f); } s=0; continue; } map[++num][0]=x; map[num][1]=y; s++; } } double dist[345]; int t[345],headd,tail=1; bool vis[345]; int main() { memset(head,-1,sizeof(head)); int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2); scan(); for(int i=1; i<=num; i++) for(int j=1; j<=num; j++) { double f=far(map[i][0],map[i][1],map[j][0],map[j][1])/1000*6; add(i,j,f); } num++; for(int i=1; i<num; i++) { double f=far(map[i][0],map[i][1],x1,y1)/1000*6; add(i,num,f); add(num,i,f); } num++; for(int i=1; i<num-1; i++) { double f=far(map[i][0],map[i][1],x2,y2)/1000*6; add(i,num,f); add(num,i,f); } t[0]=num-1; vis[num-1]=1; for(int i=1; i<=num; i++) dist[i]=2e8; dist[num-1]=0; while(headd!=tail) { int r=head[t[headd]]; printf("r=%d\n",r); while(r!=-1) { if(dist[e[r].to]>dist[t[headd]]+e[r].lon) { dist[e[r].to]=dist[t[headd]]+e[r].lon; printf("dist[e[r].to]=%f\n",dist[e[r].to]); if(!vis[e[r].to]) { vis[e[r].to]=1; t[tail++]=e[r].to; } } r=e[r].next; } headd++; } printf("%d",change(dist[num])); return 0; } // FOR C.H
POJ 2502 Subway-經過預處理的最短路