Mr. Kitayuta's Colorful Graph CodeForces - 505B

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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.

Example

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.

题意：本题很好理解,上面有图，有解释。意思是:给你三个数据：a b c，意思是a和b用c连接，如果a和d再用c链接，那么可以说b和d连接。（媒介必须相同），再给你几组数据，问每一组数据（a,b）之间有几条这样的连接方案。

思路：之前学过并查集，一维的并查集的在这里不适用（考试时，我都没想到用并查集啊）。没想到是个二维的。一看过程，发现还挺简单，代码很简单,思路很清晰（但考试时就是没想到）。

#include 
#include 
int a[110][110];
int n,m;

int find(int x,int k)
{
    if(a[x][k]==x)
        return x;
    return a[x][k]=find(a[x][k],k);
}

int merge(int x,int y,int z)
{
    int tx=find(x,z);
    int ty=find(y,z);
    if(tx!=ty)
        a[tx][z]=ty;
}

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        int x,y,z;
        for(int i=1; i<=100; i++)
            for(int j=1; j<=100; j++)
                a[i][j]=i;
        for(int i=1; i<=m; i++)
        {
            scanf("%d %d %d",&x,&y,&z);
            merge(x,y,z);
        }
        int t;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d %d",&x,&y);
            int sum=0;
            for(int i=1; i<=m; i++)
                if(find(x,i)==find(y,i))
                    sum++;
            printf("%d\n",sum);
        }
    }
    return 0;
}

考试的时候我用的floyd算法，结果做错了，现在想想，发现自己理解有错误，开三维数组，加一个四层for循环这些都没错，错在我的floyd算法错误和累加数值上，因为开的是三维的，不需要累加（因为数组较全，包含所有情况）还有写算法的时候应该注意一下不要写错。

#include 
#include 

int map[110][110][110];
int n,m;

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        int x,y,z;
        memset(map,0,sizeof(map));
        for(int i=0; i