把每个电梯口看作一个节点， 然后计算边的权值的时候处理一下， 就ok了。
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;const int MAXN = 112;
struct Edge { int v, id; };
struct node
{int t, v, id;bool operator < (const node& rhs) const{return t > rhs.t;}
};
vector<Edge> g[MAXN];
vector<int> a;
int d[MAXN], speed[10], n, k, x;void solve()
{priority_queue<node> q;REP(i, 0, MAXN) d[i] = (i == 0 ? 0 : 1e9);q.push(node{0, 0, -1});while(!q.empty()){node x = q.top(); q.pop();int u = x.v;if(x.t != d[u]) continue;REP(i, 0, g[u].size()){int v = g[u][i].v, id = g[u][i].id;int w = speed[id] * abs(u - v);if(x.id != id && x.id != -1) w += 60;if(d[v] > d[u] + w){d[v] = d[u] + w;q.push(node{d[v], v, id});}}} if(d[k] == 1e9) puts("IMPOSSIBLE");else printf("%d\n", d[k]);
}int main()
{while(~scanf("%d%d", &n, &k)){REP(i, 0, MAXN) g[i].clear();REP(i, 0, n) scanf("%d", &speed[i]);REP(i, 0, n){a.clear();scanf("%d", &x);a.push_back(x);while(getchar() != '\n'){scanf("%d", &x);a.push_back(x);}REP(r, 0, a.size())REP(j, r + 1, a.size()) {g[a[j]].push_back(Edge{a[r], i});g[a[r]].push_back(Edge{a[j], i});}}solve();}return 0;
}