CodeForces 505B Mr. Kitayuta's Colorful Graph

Mr. Kitayuta's Colorful Graph

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

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 — aibi (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.

Sample Input

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

Hint

Let's consider the first sample.

CodeForces 505B    Mr. Kitayuta's Colorful Graph 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<bits/stdc++.h>
#define maxx 105
using namespace std;
vector <int>edg[maxx][maxx];
bool vis[maxx];
int ans;
int x,y;
bool bfs(int c)
{
queue<int> q;
q.push(x);
int now;
memset(vis,,sizeof(vis));
while(!q.empty())
{
now=q.front();
if(now==y) return true;
q.pop();
for(int i=;i<edg[c][now].size();i++)
{
int then=edg[c][now][i];
if(vis[then]) continue;
vis[then]=;
q.push(then);
}
}
return false;
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
int a,b,c;
for(int i=;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
edg[c][b].push_back(a);
edg[c][a].push_back(b);
}
int q;
scanf("%d",&q);
for(int i=;i<q;i++)
{
ans=;
scanf("%d%d",&x,&y);
for(int i=;i<m;i++)
if(bfs(i+)) ans++;
printf("%d\n",ans);
}
}

这道题不难理解,用bfs,保存每种颜色的边的关联的边,然后对每种颜色的边bfs,如果能够找到符合起点终点的边,结果就加一,知道找完所有的颜色的边。

据说这题还可以用并查集做。有机会看一下。学渣现在正处于学习阶段。

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