Lift Hopping
Time Limit: 1 second
| Ted the bellhop: "I'm coming up and if there isn't a dead body by the time I get there, I'll make one myself. You!" |
A skyscraper has no more than 100 floors, numbered from 0 to 99. It has n (1<=n<=5) elevators which travel up and down at (possibly) different speeds. For each i in {1, 2,... n}, elevator number i takes Ti (1<=Ti<=100) seconds to travel between any two adjacent floors (going up or down). Elevators do not necessarily stop at every floor. What's worse, not every floor is necessarily accessible by an elevator.
You are on floor 0 and would like to get to floor k as quickly as possible. Assume that you do not need to wait to board the first elevator you step into and (for simplicity) the operation of switching an elevator on some floor always takes exactly a minute. Of course, both elevators have to stop at that floor. You are forbiden from using the staircase. No one else is in the elevator with you, so you don't have to stop if you don't want to. Calculate the minimum number of seconds required to get from floor 0 to floor k (passing floor k while inside an elevator that does not stop there does not count as "getting to floork").
Input
The input will consist of a number of test cases. Each test case will begin with two numbers, n and k, on a line. The next line will contain the numbers T1, T2,... Tn. Finally, the next n lines will contain sorted lists of integers - the first line will list the floors visited by elevator number 1, the next one will list the floors visited by elevator number 2, etc.
The input will consist of a number of test cases. Each test case will begin with two numbers, n and k, on a line. The next line will contain the numbers T1, T2,... Tn. Finally, the next n lines will contain sorted lists of integers - the first line will list the floors visited by elevator number 1, the next one will list the floors visited by elevator number 2, etc.
Output
For each test case, output one number on a line by itself - the minimum number of seconds required to get to floor k from floor 0. If it is impossible to do, print "IMPOSSIBLE" instead.| Sample Input | Sample Output |
2 30 10 5 0 1 3 5 7 9 11 13 15 20 99 4 13 15 19 20 25 30 2 30 10 1 0 5 10 12 14 20 25 30 2 4 6 8 10 12 14 22 25 28 29 3 50 10 50 100 0 10 30 40 0 20 30 0 20 50 1 1 2 0 2 4 6 8 10 | 275 285 3920 IMPOSSIBLE |
Explanation of examples
In the first example, take elevator 1 to floor 13 (130 seconds), wait 60 seconds to switch to elevator 2 and ride it to floor 30 (85 seconds) for a total of 275 seconds.In the second example, take elevator 1 to floor 10, switch to elevator 2 and ride it until floor 25. There, switch back to elevator 1 and get off at the 30'th floor. The total time is
10*10 + 60 + 15*1 + 60 + 5*10 = 285 seconds.
In example 3, take elevator 1 to floor 30, then elevator 2 to floor 20 and then elevator 3 to floor 50.
In the last example, the one elevator does not stop at floor 1.
Problemsetter: Igor Naverniouk
Alternate solutions: Stefan Pochmann, Frank Pok Man Chu
shortest path - bfs
#include<stdio.h>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<queue>
#include<sstream>
#define REP(i, b, n) for (int i = b; i < n; i++)
#define rep(i, n) REP(i, 0, n)
#define MAXN 5
#define MAXF 100
#define DBG 0
using namespace std;
int N,M,next[MAXN][MAXF],pre[MAXN][MAXF],spd[MAXN];
bool dps[MAXN][MAXF],isc[MAXN][MAXF];
vector<int>eg[MAXF];
struct Node{
Node(int a,int b,int c):ele(a),flr(b),min(c){}
bool operator < (const Node b)const{return min>b.min;}
int ele,flr,min;
};
void ans(){
int a,b,c,na,nb[2];
bool ok=0;
memset(dps,0,sizeof(dps));
priority_queue<Node>q;
rep(i,N)if(isc[i][0])dps[i][0]=1,q.push(Node(i,0,0));
while(q.size()){
a=q.top().ele,b=q.top().flr,c=q.top().min,q.pop();
if(DBG)printf("bfs %d %d %d\n",a,b,c);
if(b==M){ok=1;break;}
dps[a][b]=1;
nb[0]=next[a][b],nb[1]=pre[a][b]; //switch floor
rep(i,2)
if(nb[i]>0&&!dps[a][nb[i]]){
q.push(Node(a,nb[i],c+spd[a]*abs(nb[i]-b))); //original ele
}
rep(i,eg[b].size()){ //switch elevator
na=eg[b][i];
if(na!=a){
if(DBG)printf("switch flr: %d %d\n",na,c+60);
if(!dps[na][b])q.push(Node(na,b,c+60));}
}
}
if(ok)printf("%d\n",c);
else printf("IMPOSSIBLE\n");
}
int main(){
int a,b,c,pa;
char chs[400];
stringstream buf;
while(scanf("%d%d",&N,&M)==2){
if(DBG)printf("N M %d %d\n",N,M);
rep(i,MAXF)eg[i].clear();memset(isc,0,sizeof(isc));
memset(pre,0,sizeof(pre)),memset(next,0,sizeof(next));
rep(i,N)scanf("%d ",&spd[i]);
rep(i,N){
fgets(chs,400,stdin);pa=-1;
if(chs[strlen(chs)-1]=='\n')chs[strlen(chs)-1]='\0';
buf.clear(),buf.str(chs);
while(!buf.eof()){
buf>>a;
if(pa>=0)pre[i][a]=pa,next[i][pa]=a;
eg[a].push_back(i);isc[i][a]=1;
pa=a;
}
}
ans();
}
return 0;
}
grid of square tiles. The tiles have such size that moving forward from the center of one tile to that of the next one makes the wheel rotate exactly 72o around its own center. The effect is shown in the above figure. When the wheel is at the center of square 1, the midpoint of the periphery of its blue segment is in touch with the ground. But when the wheel moves forward to the center of the next square (square 2) the midpoint of its white segment touches the ground.
, 
) giving the dimensions of the grid. Then follows the description of the grid in M lines of N characters each. The character `#' will indicate a blocked square, all other squares are free. The starting location of the cyclist is marked by `S' and the target is marked by `T'. The input terminates with two zeros for M and N.
, 
and 
where C indicates the number of crossings (crossings are numbered using distinct integers ranging from 1 to C), S represents the number of streets and Q is the number of queries.
) is d decibels.
