B. Mr. Kitayuta's Colorful Graph
1 second
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.
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.
For each query, print the answer in a separate line.
4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4
2
1
0
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
1
1
1
1
2
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.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <vector>
using namespace std;
const double EXP=1e-;
const double PI=acos(-1.0);
const int INF=0x7fffffff;
const int MS=; int fa[MS][MS];
void init()
{
for(int i=;i<MS;i++)
for(int j=;j<MS;j++)
fa[i][j]=j;
}
int find(int c,int a)
{
if(fa[c][a]==a)
return a;
return fa[c][a]=find(c,fa[c][a]);
}
void make_union(int c,int a,int b)
{
a=find(c,a);
b=find(c,b);
if(a==b)
return ;
fa[c][b]=fa[c][a];
}
int main()
{
int n,m,q,a,b,c;
init();
cin>>n>>m;
for(int i=;i<m;i++)
{
cin>>a>>b>>c;
make_union(c,a,b);
}
cin>>q;
while(q--)
{
int ans=;
cin>>a>>b;
for(c=;c<=m;c++)
if(find(c,a)==find(c,b))
ans++;
cout<<ans<<endl;
}
return ;
}
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