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.
給定n個點和m條邊,點編號從1~n,每條邊有:起點,終點和顏色
問給定任意兩點,求連接這兩點所有顏色的數量.(如:紅色要連接兩個點,那么這兩點間必須存在一條顏色為紅色的路徑,即兩點連通於紅色)
求連通性很容易想到並查集.將每種顏色看成一個並查集,要求兩點之間所有顏色時只需遍歷所有顏色,看有多少種顏色能連通即可.
#include <iostream> using namespace std; class bingchaji { public: bingchaji() { init(); } int father[200]; void init() { for(int i=0;i<200;i++) father[i]=i; } void unit(int u,int v) { father[getfather(u)]=father[getfather(v)]; } int getfather(int v) { if(father[v]==v) return v; return father[v]=getfather(father[v]); } bool same(int u,int v) { return getfather(u)==getfather(v); } }color[200]; int main() { int n,m; cin>>n>>m; for(int i=1;i<=m;i++) { int u,v,c; cin>>u>>v>>c; color[c].unit(u,v); } int k; cin>>k; for(int i=1;i<=k;i++) { int u,v; int res=0; cin>>u>>v; for(int j=1;j<=m;j++) { if(color[j].same(u,v)) res++; } cout<<res<<endl; } return 0; }