CodeForces - 505B-Mr. Kitayuta's Colorful Graph(暴力)

Mr. Kitayuta has just bought an undirected graph consisting of n vertices and medges. 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.

Examples

Input

4 51 2 11 2 22 3 12 3 32 4 331 23 41 4

Output

Input

5 71 5 12 5 13 5 14 5 11 2 22 3 23 4 251 55 12 51 51 4

Output

Note

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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
struct node
{
    int v;
    int val;
};
int ans=0;
int vis[105];
vector<node>vec[105];
void bfs(int x,int y)
{
    node start;
    for(int j=1;j<=100;j++)
    {
    queue<node>q;
    memset(vis,0,sizeof(vis));
    vis[x]=1;
    for(int t=0;t<vec[x].size();t++)
    {
        if(vec[x][t].val==j&&vis[vec[x][t].v]==0)
        {
            q.push(vec[x][t]);
            vis[vec[x][t].v]=1;
        }
    }
    while(!q.empty())
    {
        node now=q.front();
        q.pop();
        if(now.v==y)
        {
            ans++;
        }
        for(int t=0;t<vec[now.v].size();t++)
        {
            if(vec[now.v][t].val==j&&vis[vec[now.v][t].v]==0)
            {
                vis[vec[now.v][t].v]=1;
                q.push(vec[now.v][t]);
            }
        }
    }
    }
}
int main()
{
    int n,m;
    cin>>n>>m;

    int a,b,c;
    set<int>s;
    for(int t=0;t<m;t++)
    {
        scanf("%d%d%d",&a,&b,&c);
        s.insert(c);
        node st;
        st.v=b;
        st.val=c;
        vec[a].push_back(st);
        st.v=a;
        st.val=c;
        vec[b].push_back(st);
    }
    int q;
    cin>>q;
    int uu,vv;
    for(int t=0;t<q;t++)
    {
        scanf("%d%d",&uu,&vv);
        ans=0;
        bfs(uu,vv);
        printf("%d\n",ans);
    }
    
    return 0;
}

来源：https://www.cnblogs.com/Staceyacm/p/10858718.html

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