505B. Mr. Kitayuta's Colorful Graph（拆分图+并查集）

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本文介绍了一个基于图论的问题，即如何在一个拥有多种颜色边的无向图中，针对特定顶点对查询连接它们的不同颜色数量。通过使用并查集算法进行图的拆分，文章详细解释了处理这一问题的具体步骤及其实现代码。

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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 c_i, connecting vertex a_i and b_i.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers — u_i and v_i.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex u_i and vertex v_i directly or indirectly.

Input

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 — a_i, b_i (1 ≤ a_i < b_i ≤ n) and c_i (1 ≤ c_i ≤ 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, (a_i, b_i, c_i) ≠ (a_j, b_j, c_j).

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 — u_i and v_i (1 ≤ u_i, v_i ≤ n). It is guaranteed that u_i ≠ v_i.

Output

For each query, print the answer in a separate line.

Sample test(s)

Input

Output

2
1
0

Input

Output

Note

Let's consider the first sample.

$\text{[math]}$ The figure above shows 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.

图的拆分加并查集。

根据颜色将图进行拆分，然后并查集查找连通的数量

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int pre[110][110];

int find(int *pre,int x){
while(x!=pre[x]){
  x = pre[x];
}
return x;
}
int main(){
int n,m,u,v,c;
int pre_u,pre_v;
while(~scanf("%d%d",&n,&m)){
  for(int i=1;i<=m;i++){
   for(int j = 1;j<=n;j++)
   pre[i][j] = j;
  }

  for(int i =1;i<=m;i++){
   scanf("%d%d%d",&u,&v,&c);
    pre_u = find(pre[c],u);
    pre_v = find(pre[c],v);
   if(pre_u!=pre_v)pre[c][pre_u] = pre_v;
  }
  int num;
  scanf("%d",&num);
  while(num--){
   int s = 0;
   scanf("%d%d",&u,&v);
   for(int i=1;i<=m;i++){
    pre_u = find(pre[i],u);
    pre_v = find(pre[i],v);
    //printf("%d %d\n",pre_u,pre_v);
    if(pre_u == pre_v)s++;
   }
   printf("%d\n",s);
  }
}
}