http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1742
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!" |
Robert Rodriguez, "Four Rooms."
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 floor k").
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.
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
題意：
有若干部電梯，給出每部電梯通過一層的時間和能停的樓層，換乘電梯要用60s，電梯可上可下，求從0層到目標層的最短時間。
分析：
最短路。假設目前在i電梯j層，那麼能更新的狀態有兩種：i電梯的其他樓層和j樓層的其他電梯。這也是建圖的依據，感覺這題不顯式地建圖會方便些。
/*
*
* Author : fcbruce
*
* Date : 2014-09-03 16:31:53
*
*/
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#ifdef _WIN32
#define lld "%I64d"
#else
#define lld "%lld"
#endif
#define maxm
#define maxn 101
using namespace std;
int n,obj;
int v[5];
vector <int> stop[5];
int G[5][maxn],d[5][maxn];
pair<int,int> q[maxn<<8];
bool inq[5][maxn];
void SPFA()
{
memset(d,0x3f,sizeof d);
memset(inq,0,sizeof inq);
int f=0,r=-1;
for (int i=0;i<n;i++)
if (G[i][0])
{
q[++r]=make_pair(i,0);
d[i][0]=0;
}
while (f<=r)
{
pair<int,int> x=q[f++];
inq[x.first][x.second]=false;
for (int u:stop[x.first])
{
if (d[x.first][u]>d[x.first][x.second]+abs(x.second-u)*v[x.first])
{
d[x.first][u]=d[x.first][x.second]+abs(x.second-u)*v[x.first];
if (!inq[x.first][u])
{
inq[x.first][u]=true;
q[++r]=make_pair(x.first,u);
}
}
}
for (int i=0;i<n;i++)
{
if (x.first==i || G[i][x.second]==false) continue;
if (d[i][x.second]>d[x.first][x.second]+60)
{
d[i][x.second]=d[x.first][x.second]+60;
if (!inq[i][x.second])
{
inq[i][x.second]=true;
q[++r]=make_pair(i,x.second);
}
}
}
}
}
int main()
{
#ifdef FCBRUCE
freopen("/home/fcbruce/code/t","r",stdin);
#endif // FCBRUCE
char str[233];
while (~scanf( "%d%d",&n,&obj))
{
memset(G,0,sizeof G);
for (int i=0;i<n;i++)
scanf( "%d",v+i);
getchar();
for (int i=0;i<n;i++)
{
gets(str);
stringstream s(str);
int x;
stop[i].clear();
while (s>>x)
{
stop[i].push_back(x);
G[i][x]=true;
}
}
SPFA();
int MIN=INF;
for (int i=0;i<n;i++)
if (G[i][obj]) MIN=min(MIN,d[i][obj]);
if (MIN==INF)
puts( "IMPOSSIBLE");
else
printf( "%d\n",MIN);
}
return 0;
}
