1 /*
2 题意：给出n个地点，每个地点的油价为pi每单位，给出m条边，每条长度为d，行走d距离的路需
3 要d单位的油；给出一辆车的油箱容量以及起始点:s,e；问s到e最少要耗费多少钱
4
5 题解：BFS+优先队列
6 这个搜索方式比较巧妙：状态为对于当前点是走还是油+1，这样就算想到了一般人也不太敢做，
7 毕竟看起来复杂度还是蛮高的；其中加了一个优先队列，作用是通过类似dijkstra的贪心的方法
8 保证求出的是最值，通过vis[][]记录状态即可去除重复。
9 */
10 #include <cstdio>
11 #include <cstring>
12 #include <queue>
13
14 using namespace std;
15
16 const int MAXV = 1005;
17 const int MAXE = 30005;
18
19 struct edge
20 {
21 int v,w,next;
22 }E[MAXE];
23
24 int EN;
25 int head[MAXV];
26
27 void insert(int u, int v, int w)
28 {
29 E[EN].v = v;
30 E[EN].w = w;
31 E[EN].next = head[u];
32 head[u] = EN++;
33 E[EN].v = u;
34 E[EN].w = w;
35 E[EN].next = head[v];
36 head[v] = EN++;
37 }
38
39 int p[MAXV];
40
41 struct node
42 {
43 int cost,v,oil;
44 bool operator<(const node &t)const
45 {
46 return cost > t.cost;
47 }
48 };
49
50 bool vis[105][MAXV];
51
52 int bfs(int c, int s, int e)
53 {
54 memset(vis,false,sizeof(vis));
55 priority_queue<node> Q;
56 node start;
57 start.cost = 0;
58 start.oil = 0;
59 start.v = s;
60 Q.push(start);
61 while (!Q.empty())
62 {
63 node now = Q.top();
64 Q.pop();
65 if (now.v == e)
66 return now.cost;
67
68 vis[now.oil][now.v] = true;
69
70 if (now.oil < c && !vis[now.oil+1][now.v])
71 {
72 now.oil++;
73 now.cost += p[now.v];
74 Q.push(now);
75 now.oil--;
76 now.cost -= p[now.v];
77 }
78
79 for(int i=head[now.v]; i!=-1; i=E[i].next)
80 {
81 if (now.oil >= E[i].w && !vis[now.oil-E[i].w][E[i].v])
82 {
83 node tmp = now;
84 tmp.oil -= E[i].w;
85 tmp.v = E[i].v;
86 Q.push(tmp);
87 }
88 }
89 }
90 return -1;
91 }
92
93 int main(void)
94 {
95 int n,m;
96 while (~scanf("%d%d",&n,&m))
97 {
98 EN = 0;
99 memset(head,-1,sizeof(head));
100 for(int i=0; i<n; i++)
101 scanf("%d",&p[i]);
102 for(int i=0; i<m; i++)
103 {
104 int u,v,d;
105 scanf("%d%d%d",&u,&v,&d);
106 insert(u,v,d);
107 }
108
109 int q;
110 scanf("%d",&q);
111 while (q--)
112 {
113 int s,e,c;
114 scanf("%d%d%d",&c,&s,&e);
115 int ans = bfs(c,s,e);
116 if (ans < 0)
117 printf("impossible\n");
118 else
119 printf("%d\n",ans);
120 }
121 }
122 return 0;
123 }
