GRE Words Once More!

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 205    Accepted Submission(s): 32

Problem Description

Now Matt is preparing for the Graduate Record Examinations as Coach Pang did in 2013 and George did in 2011.

Thanks to modern techniques, Matt uses automata instead of old-fasioned vocabulary books.

The automata used by Matt is a directed acyclic graph (DAG) with N vertices and M edges. The vertices are conveniently numbered by 1, 2, . . . , N . Each edge is labeled with an integer. Additionally, some vertices are marked as special.

A GRE word is obtained by concatenating the labels on the path from vertex 1 to a special vertex.

Now, Matt has Q questions. The i-th question is asking for the length of ki-th smallest words among all the GRE words he can obtain in lexicographical order.

Input

The first line contains only one integer T , which indicates the number of test cases.

For each test case, the first line contains three integers N, M, Q (2 ≤ N ≤ 10⁵, 0 ≤ M ≤ 10⁵, 1 ≤ Q ≤ 10⁵).

The second line contains N - 1 integers s₂, . . . , s_n. If the i-th vertex is special, then s_i = 1. Otherwise, s_i = 0. Vertex 1 is never special.

Each of the following M lines contains three integers a_i, b_i, c_i denoting an edge from vertex ai to vertex b_i labeled with c_i (1 ≤ a_i, b_i ≤ N, 1 ≤ c_i ≤ 10⁹). For each vertex v, all outgoing edges are labeled with distinct integers.

Each of the following Q lines contains the integer ki (1 ≤ k_i ≤ 10⁸) of the i-th question.

Output

For each test case, output “Case #x:” in the frirst line, where x is the case number (starting from 1).

Then, for each question, output the length of the word in one line. If the word does not exist, output “-1” (without quotes) instead.

Sample Input

1
3 3 4
1 1
1 2 1
1 3 12
2 3 3
1
2
3
4

Sample Output

Case #1:
1
2
1
-1

Hint

There are 3 GRE words in total (sorted in lexicographical order):
1. (1)
2. (1, 3)
3. (12)

　　这道题不是很难，需要注意清空数组。

　　思路是预处理答案，DFS时用手写栈防爆栈，有个必要的优化，就是扫过后答案是可以重复利用的。

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <vector>
 6 using namespace std;
 7 const int N=200010,M=100000000;
 8 vector<pair<int,int> >g[N];
 9 int ans[M+10],f[N],be[N],ed[N],tot;
10 int st[N],dep[N],vis[N],mem[N],top;
11 int T,cas=0,q,n,m,Q;
12 int main(){
13     scanf("%d",&T);
14     while(T--){
15         scanf("%d%d%d",&n,&m,&Q);tot=0;
16         for(int i=2;i<=n;i++)scanf("%d",&f[i]);
17         for(int i=1,a,b,v;i<=m;i++){
18             scanf("%d%d%d",&a,&b,&v);
19             g[a].push_back(make_pair(v,b));
20         }
21         for(int i=1;i<=n;i++)
22             sort(g[i].begin(),g[i].end());
23         st[top=1]=1;dep[top]=0;
24         memset(vis,0,sizeof(vis));
25         memset(be,0,sizeof(be));
26         memset(ed,0,sizeof(ed));
27         while(top){
28             int x=st[top],d=dep[top];
29             if(vis[top]){
30                 if(!ed[x])ed[x]=tot;
31                 vis[top]=0;top-=1;
32                 continue;
33             }
34             vis[top]=1;
35             if(be[x]){
36                 int depth=-mem[x]+d;
37                 for(int i=be[x];i<=ed[x];i++){
38                     ans[++tot]=ans[i]+depth;
39                     if(tot>=M)break;
40                 }if(tot>=M)break;
41                 continue;
42             }
43             be[x]=tot+1;mem[x]=d;
44             if(f[x])ans[++tot]=d;
45             if(tot>=M)break;
46             for(int i=g[x].size()-1;~i;i--){
47                 st[++top]=g[x][i].second;
48                 dep[top]=d+1;
49             }
50         }
51         printf("Case #%d:\n",++cas);
52         while(Q--){
53             scanf("%d",&q);
54             if(q>tot)printf("-1\n");
55             else printf("%d\n",ans[q]);
56         }
57         for(int i=1;i<=n;i++)g[i].clear();
58     }
59     return 0;
60 }