CodeForces 505B Mr. Kitayuta's Colorful Graph

Mr. Kitayuta's Colorful Graph

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 505B

Description

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 Input

Input

4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4

Output

2
1
0

Input

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

Output

1
1
1
1
2

Hint

Let's consider the first sample.

 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.

#include<bits/stdc++.h>
#define maxx 105
using namespace std;
vector <int>edg[maxx][maxx];
bool vis[maxx];
int ans;
int x,y;
bool bfs(int c)
{
    queue<int> q;
    q.push(x);
    int now;
    memset(vis,,sizeof(vis));
    while(!q.empty())
    {
        now=q.front();
        if(now==y) return true;
        q.pop();
        for(int i=;i<edg[c][now].size();i++)
        {
            int then=edg[c][now][i];
            if(vis[then]) continue;
            vis[then]=;
            q.push(then);
        }
    }
    return false;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    int a,b,c;
    for(int i=;i<m;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        edg[c][b].push_back(a);
        edg[c][a].push_back(b);
    }
    int q;
    scanf("%d",&q);
    for(int i=;i<q;i++)
    {
        ans=;
        scanf("%d%d",&x,&y);
        for(int i=;i<m;i++)
            if(bfs(i+)) ans++;
         printf("%d\n",ans);
    }
}

这道题不难理解，用bfs，保存每种颜色的边的关联的边，然后对每种颜色的边bfs，如果能够找到符合起点终点的边，结果就加一，知道找完所有的颜色的边。

据说这题还可以用并查集做。有机会看一下。学渣现在正处于学习阶段。