Mr. Kitayuta's Colorful Graph(二维并查集,弱化版)

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附加强版链接(待做):Mr. Kitayuta's Colorful Graph - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
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
Copy
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
Copy
2
1
0
input
Copy
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
Copy
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.
复制代码
//https://www.luogu.com.cn/problem/CF506D
//由于数据较小,运用二维并查集,第二维用于储存颜色
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int p[N][N],res,n,m,num;
int find(int x,int color)
{
    if(x!=p[x][color]) p[x][color]=find(p[x][color],color);
    return p[x][color];
}
int main()
{
    std::ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++) p[i][j]=i;
    for(int i=1;i<=m;i++){
        int a,b,c;
        cin>>a>>b>>c;
        p[find(a,c)][c]=find(b,c);
    }
    cin>>num;
    while(num--){
        int u,v,res=0;
        cin>>u>>v;
        for(int i=0;i<=m;i++)
            if(find(u,i)==find(v,i)) res++;
        cout<<res<<endl;
    }
    return 0;
}
复制代码

 


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本文作者：Sakurajimamai
本文链接：https://www.cnblogs.com/o-Sakurajimamai-o/p/17485327.html
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