Codeforces 505B Mr Kitayuta ' s colorful Graph

Mr Kitayuta ' s colorful GraphTime limit:1000MS Memory Limit:262144KB 64bit IO Format:%i64d &%i6 4u SubmitStatusPracticecodeforces 505B

Description

Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the graph is numbered from 1 to n. Each edge, namely Edge I, have a color c_{i, connecting vertex a i and bi.}

Mr. Kitayuta wants you to process the following q queries.

in the I-th query, he gives you and integers- u_{i and vI.}

Find the number of the colors that satisfy the following condition:the edges of this color connect vertex u_{i and vertex vi directly or indirectly.}

Input

The first line of the input contains space-separated, integers- n and m (2≤ n ≤100, 1≤ m ≤100), denoting the number of the vertices and the number of the edges, respectively.

The nextmLines contain space-separated three integers-a_{I,b i   (1≤ a I < b i ≤ n ) and  c  Sub class= "Lower-index" > i   (1≤ C i ≤ m ). Note that there can is multiple edges between the vertices. However, there is no multiple edges of the same color between II vertices, that's 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 II integers- u_{i and V i (1≤ ui, vi ≤ n). It's guaranteed that ui ≠ vi.}

Output

For each query, print the answer-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.

The figure above shows the first sample.

• Vertex 1 and Vertex 2 is connected by color 1 and 2.
• Vertex 3 and Vertex 4 is connected by Color 3.
• Vertex 1 and Vertex 4 is not connected by any single color.

#include <bits/stdc++.h>#defineMaxx 105using namespaceStd;vector<int>Edg[maxx][maxx];BOOLVis[maxx];intans;intx, y;BOOLBFsintc) {Queue<int>Q;    Q.push (x); intNow ; memset (Vis,0,sizeof(VIS));  while(!Q.empty ()) { Now=Q.front (); if(now==y)return true;        Q.pop ();  for(intI=0; I<edg[c][now].size (); i++)        {            intthen=Edg[c][now][i]; if(Vis[then])Continue; Vis[then]=1;        Q.push (then); }    }    return false;}intMain () {intn,m; scanf ("%d%d",&n,&m); intA,b,c;  for(intI=0; i<m;i++) {scanf ("%d%d%d",&a,&b,&c);        Edg[c][b].push_back (a);    Edg[c][a].push_back (b); }    intQ; scanf ("%d",&q);  for(intI=0; i<q;i++) {ans=0; scanf ("%d%d",&x,&y);  for(intI=0; i<m;i++)            if(BFS (i+1)) ans++; printf ("%d\n", ans); }}

View Code

This problem is not difficult to understand, with BFS, save each color edge of the associated edge, and then for each color of the edge of BFS, if you can find the end of the edge to match the beginning, the result is added one, know to find out all the color of the side.

It is said that the problem can also be used and check the collection. Have a chance to take a look. The dregs are now at the learning stage.

Codeforces 505B Mr Kitayuta ' s colorful Graph