Portal

Problem Description

ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.

Input

There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).

Output

Output the answer to each query on a separate line.

Sample Input

10 10 10

7 2 1

6 8 3

4 5 8

5 8 2

2 8 9

6 4 5

2 1 5

8 10 5

7 3 7

7 8 8

Sample Output

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cmath>
  4 #include <iostream>
  5 #include <algorithm>
  6 using namespace std;
  7 #define N 10005
  8 #define M 50005
  9 /*
 10 因为询问比较多，所以需要离线
 11 在线的意思就是每一个询问单独处理复杂度O（多少多少)，
 12 离线是指将所有的可能的询问先一次都处理出来，
 13 最后对于每个询问O(1)回答
 14 
 15 对于每个询问有一个长度L，问有多少条路径的长度<=L
 16 而且该路径的长度是T，T是从u到v上最长的边
 17 只要求得有多少个点对使得点对之间的最大的边小于L即可。
 18 
 19 先从小到大排序一遍询问的边的长度L，从小到大枚举u->v之间的边的长度
 20 如果两个集合没有联通，那么联通之后路径的条数为sum[x]*sum[y]
 21 因为长的L必定包含了短的L的答案，所以要累加起来
 22 */
 23 int fa[N],sum[N];
 24 
 25 struct node1
 26 {
 27     int id,ans,l;
 28 }query[N];
 29 
 30 struct node2
 31 {
 32     int u,v,len;
 33 }edge[M];
 34 
 35 bool cmp1(node2 a,node2 b)
 36 {
 37     return a.len<b.len;
 38 }
 39 
 40 bool cmp2(node1 a,node1 b)
 41 {
 42     return a.id<b.id;
 43 }
 44 
 45 bool cmp3(node1 a,node1 b)
 46 {
 47     return a.l<b.l;
 48 }
 49 
 50 int Find(int x)
 51 {
 52     if(x==fa[x]) return x;
 53     return fa[x]=Find(fa[x]);
 54 }
 55 
 56 int Merge(int x,int y)
 57 {
 58     int fx=Find(x),fy=Find(y);
 59     if(fx==fy) return 0;
 60     int tmp;
 61     if(fx<fy){
 62         fa[fy]=fx;
 63         tmp=sum[fx]*sum[fy];
 64         sum[fx]+=sum[fy];
 65     }
 66     else{
 67         fa[fx]=fy;
 68         tmp=sum[fx]*sum[fy];
 69         sum[fy]+=sum[fx];
 70     }
 71     return tmp;
 72 }
 73 
 74 int main()
 75 {
 76     int n,m,q;
 77     while(~scanf("%d%d%d",&n,&m,&q)){
 78         for(int i=1;i<=n;i++){
 79             fa[i]=i;
 80             sum[i]=1;
 81         }
 82         for(int i=0;i<m;i++){
 83             scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].len);
 84         }
 85         for(int i=0;i<q;i++){
 86             scanf("%d",&query[i].l);
 87             query[i].id=i;
 88             query[i].ans=0;
 89         }
 90         sort(edge,edge+m,cmp1);
 91         sort(query,query+q,cmp3);
 92         int cnt=0;
 93         for(int i=0;i<q;i++){
 94             while(edge[cnt].len<=query[i].l&&cnt<m){
 95                 int x=edge[cnt].u;
 96                 int y=edge[cnt].v;
 97                 int fx=Find(x);
 98                 int fy=Find(y);
 99                 if(fx==fy) cnt++;
100                 else{
101                     query[i].ans+=Merge(x,y);
102                     cnt++;
103                 }
104             }
105             if(i>0) query[i].ans+=query[i-1].ans;
106         }
107         sort(query,query+q,cmp2);
108         for(int i=0;i<q;i++){
109             printf("%d
",query[i].ans);
110         }
111     }
112     return 0;
113 }