Lift Hopping

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.
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.
Sample Input
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
一座大楼0到99层，最多有5个电梯，每个电梯有自己的速度，和自己独特的停靠楼层，换电梯需要额外的60s，问到达目标楼层的最短时间。
图实在是太明显了，直接100层而已，然后直接用迪杰斯特拉一个一个找。
一开始赋值表达式写了比较之前的电梯与之后电梯是否一致……后来发现根本不需要比较，因为如果之前就是这部电梯，那么到达的那个点一定在之前搜索过了，而且是不加60s的情况，所以就算表达式中加60s与之前比较，也会被过滤掉。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
struct Node{
    int v;
    int leap[105];
}node[10];
int vis[105], d[105];
int main()
{
    int i, j, m, n, ans, t, k;
    while (scanf("%d%d", &n, &k) != EOF)
    {
        for (i = 1; i <= n; i++)
        {
            cin >> node[i].v;
            for (j = 0; j <= 100; j++)
                node[i].leap[j] = 0;
        }
        int cnt = 1;
        while (cnt <= n)
        {
            int x;
            char c;
            c = 'a';
            while (c != '\n'){
                cin >> x;
                node[cnt].leap[x] = 1;
                c = getchar();
            }
            cnt++;
        }
        memset(vis, 0, sizeof(vis));
        for (i = 1; i <= 100; i++)
            d[i] = inf;
        d[0] = 0;
        vis[0] = 1;
        for (i = 1; i <= n; i++)
        {
            if (node[i].leap[0])
            {
                for (j = 1; j <= 100; j++)
                if (node[i].leap[j])
                {
                    d[j] = min(d[j], node[i].v*j);
                }
            }
        }
        for (i = 1; i <= 100; i++)
        {
            int tempx, tempm;
            tempx = tempm = inf;
            for (j = 1; j <= 100; j++)
            {
                if (!vis[j] && d[j] <= tempm)
                {
                    tempm = d[j];
                    tempx = j;
                }
            }
            if (tempx == inf)break;
            vis[tempx] = 1;
            for (j = 1; j <= 100; j++)
            {
                int k;
                for (k = 1; k <= n; k++)
                {
                    if (node[k].leap[j] && node[k].leap[tempx])
                    {
                        d[j] = min(d[j], d[tempx] + node[k].v*abs(tempx - j) + 60);
                    }
                }
            }
        }
        if (d[k] == inf)
            cout << "IMPOSSIBLE" << endl;
        else
            cout << d[k] << endl;
    }
    return 0;
}