Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color ci, connecting vertex ai and bi.
Mr. Kitayuta wants you to process the following q queries.
In the i-th query, he gives you two integers — ui and vi.
Find the number of the colors that satisfy the following condition: the edges of that color connect vertex ui and vertex vi directly or indirectly.
The first line of the input contains space-separated two integers — n and m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100), denoting the number of the vertices and the number of the edges, respectively.
The next m lines contain space-separated three integers — ai, bi (1 ≤ ai < bi ≤ n) and ci (1 ≤ ci ≤ m). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if i ≠ j, (ai, bi, ci) ≠ (aj, bj, cj).
The next line contains a integer — q (1 ≤ q ≤ 100), denoting the number of the queries.
Then follows q lines, containing space-separated two integers — ui and vi (1 ≤ ui, vi ≤ n). It is guaranteed that ui ≠ vi.
For each query, print the answer in a separate line.
4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4
2
1
0
5 7
1 5 1
2 5 1
3 5 1
4 5 1
1 2 2
2 3 2
3 4 2
5
1 5
5 1
2 5
1 5
1 4
1
1
1
1
2
Let's consider the first sample.
- Vertex 1 and vertex 2 are connected by color 1 and 2.
- Vertex 3 and vertex 4 are connected by color 3.
- Vertex 1 and vertex 4 are not connected by any single color.
//https://www.luogu.com.cn/problem/CF506D //由于数据较小,运用二维并查集,第二维用于储存颜色 #include<bits/stdc++.h> using namespace std; const int N=1e5+10; int p[N][N],res,n,m,num; int find(int x,int color) { if(x!=p[x][color]) p[x][color]=find(p[x][color],color); return p[x][color]; } int main() { std::ios::sync_with_stdio(false); cin>>n>>m; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) p[i][j]=i; for(int i=1;i<=m;i++){ int a,b,c; cin>>a>>b>>c; p[find(a,c)][c]=find(b,c); } cin>>num; while(num--){ int u,v,res=0; cin>>u>>v; for(int i=0;i<=m;i++) if(find(u,i)==find(v,i)) res++; cout<<res<<endl; } return 0; }