codeforces 505B Mr. Kitayuta's Colorful Graph(水题)

转载请注明出处： http://www.cnblogs.com/fraud/           ——by fraud

Mr. Kitayuta's Colorful Graph

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

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

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.

并查集，暴力都可以过，大水题

#include <iostream>
using namespace std;
#define MAXN 110
int pa[110][110],ra[110][110];
void init()
{
    for(int i=0;i<MAXN;i++)
    {
        for(int j=0;j<MAXN;j++){
            pa[i][j]=j;
            ra[i][j]=0;
        }
    }
}
int find(int x,int c){
    if(pa[c][x]!=x)pa[c][x]=find(pa[c][x],c);
    return pa[c][x];
}
int unite(int x,int y,int c){
    x=find(x,c);
    y=find(y,c);
    if(x==y)return 0;
    if(ra[c][x]<ra[c][y])
    {
        pa[c][x]=y;
    }else{
        pa[c][y]=x;
        if(ra[c][x]==ra[c][y])ra[c][x]++;
    }
    return 1;
}
bool same(int x,int y,int c){
    return find(x,c)==find(y,c);
}

int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    init();
    cin>>n>>m;
    int u,v,c;
    for(int i=0;i<m;i++){
        cin>>u>>v>>c;
        u--,v--,c--;
        unite(u,v,c);
    }
    int q;
    cin>>q;
    for(int i=0;i<q;i++){
        cin>>u>>v;
        u--;v--;
        int ans=0;
        for(int i=0;i<m;i++)
        {
            if(same(u,v,i))ans++;
        }
        cout<<ans<<endl;
    }

    return 0;
}