Mr. Kitayuta's Colorful Graph（二维并查集）-优快云博客

本文介绍了一道关于并查集的应用题，通过引入颜色属性来判断两点间是否存在特定颜色的连接路径，并提供了解题思路及代码实现。

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.

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 — 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.

Output

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

Examples

Input

Output

2
1
0

Input

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.

题意：有n个点和m条线，a点和b点之间用c颜色的线相连，在给定的数据连完后，进行q次查询，问从x到y点之间有多少种连线方式。

思路：这个题很像以前的并查集，不过多了一些条件限制，因此我们就要将并查集升级一下了。平常并查集f[i]只是代表这个点的祖宗。由于多了颜色的限制，我们就要把一维变成二维，第二维是颜色（f[i][j] 第i个节点，第j种颜色），查询的时候遍历每种颜色即可。

代码如下：

#include<cstdio>
#include<cstring>
int f[110][110];//f[i][j]  第i个节点，第j种颜色
void init() //初始化并查集
{
    for(int i=0;i<=100;i++) //点
        for(int j=0;j<=100;j++) //颜色
            f[i][j]=i;
}
int getf(int u,int v)//u是点，v是颜色
{
    if(f[u][v]==u)
        return u;
    return f[u][v]=getf(f[u][v],v);
}
void merge(int x,int y,int z)
{
    int t1=getf(x,z);
    int t2=getf(y,z);
    if(t1!=t2)   //根不同
        f[t2][z]=t1;
}
int main()
{
    int n,m,q;
    int x,y,z;
    init();
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        merge(x,y,z); //并查集
    }
    scanf("%d",&q);
    for(int i=0;i<q;i++)
    {
        scanf("%d%d",&x,&y);
        int ans=0;
        for(int j=1;j<=m;j++)//遍历每一种颜色
        {
            int t1=getf(x,j);
            int t2=getf(y,j);
            if(t1==t2)  //在一个集合
                ans++;
        }
        printf("%d\n",ans);
    }
}