HOJ C - Portal——并查集+离线

C - Portal

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status

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
10
6
1
5
9
1
8
2
7
6

Sample Output

36
13
1
13
36
1
36
2
16
13

#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int N= 10003;
const int M= 50002;
struct  Node
{
    int u,v;
    int len;
} edge[M];
bool cmp(Node a,Node b)
{
    return a.len<=b.len;
}
int fa[N];
int sum[N];//以该节点所在的集合点的数量
int n,m,q;
void init()
{
    for(int i=1; i<=n; i++)
        fa[i]=i,sum[i]=1;
}
int findx(int x)
{
    return fa[x]=fa[x]==x?x:findx(fa[x]);
}
int Union(int a,int b)
{
    int aa=findx(a);
    int bb=findx(b);
    if(aa==bb)
        return 0;
    else if(fa[aa]=1)
                que[i].ans+=que[i-1].ans;
        }
        sort(que,que+q,cmp2);
        for(int i=0; i