CodeForces - 505B Mr. Kitayuta's Colorful Graph（二维并查集）

题目链接：http://codeforces.com/problemset/problem/505/B点击打开链接

B. Mr. Kitayuta's Colorful Graph

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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. 

数据量很小  可以用并查集 主要就是构造每个颜色的并查集 这样才能区别每个地点是否与上一个地点相连

#include 
using namespace std;
#define maxn 200
int pre[maxn][maxn];
int findx(int col,int x)
{
    int r=x;
    while(pre[col][r]!=r)
    {
        r=pre[col][r];
    }
    int i=x;int j;
    while(pre[col][i]!=r)
    {
        j=pre[col][i];
        pre[col][i]=r;
        i=j;
    }
    return r;
}

void join (int col,int x,int y)
{
    int p1=findx(col,x);
    int p2=findx(col,y);
    if(p1!=p2)
    {
        pre[col][p2]=p1;
    }
}
int judge(int col,int x,int y)
{
    if(findx(col,x)!=findx(col,y))
        return 0;
    else
        return 1;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i