題干：

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.

解題報告：

? ? ? ?

ac代碼1：（并查集）（15ms，0.1MB）

? ?原來的代碼加上了ra數組，至今不明白是干啥的。

#include <iostream> using namespace std; #define MAXN 110 int pa[110][110],ra[110][110]; void init() {for(int i=0; i<MAXN; i++) {for(int j=0; j<MAXN; j++) {pa[i][j]=j;ra[i][j]=0;}} } int getf(int x,int c) {if(pa[c][x]!=x)pa[c][x]=getf(pa[c][x],c);return pa[c][x]; } void merge(int x,int y,int c) {int t1=getf(x,c);int t2=getf(y,c);if(t1==t2) return;else {pa[c][t2]=pa[c][t1];} // if(ra[c][t1]<ra[c][t2]) { // pa[c][t1]=t2; // } else { // pa[c][t2]=t1; // if(ra[c][t1]==ra[c][t2])ra[c][t1]++; // } } bool same(int x,int y,int c) {return getf(x,c)==getf(y,c); }int main() {ios::sync_with_stdio(false);int n,m;init();cin>>n>>m;int u,v,c;for(int i=0; i<m; i++) {cin>>u>>v>>c;u--,v--,c--;merge(u,v,c);}int q;cin>>q;for(int i=0; i<q; i++) {cin>>u>>v;u--;v--; int ans=0;for(int i=0; i<m; i++) {if(same(u,v,i))ans++;}cout<<ans<<endl;}return 0; }

ac代碼2：（廣搜bfs）（31ms，0.3MB）

#include <iostream> #include <cstring> #include <cstdio> #include <vector> #include <algorithm> #include <queue> #define MAX 107using namespace std;int n,m,c,x,y,q; vector<int> e[MAX][MAX]; bool vis[MAX];void add ( int u , int v , int w ) {e[w][u].push_back ( v );e[w][v].push_back ( u ); }bool bfs ( int c ) {memset ( vis , 0 , sizeof ( vis ) );queue<int> q;q.push ( x );while ( !q.empty() ){int u = q.front();if ( u == y ) return true;q.pop();for ( int i = 0 ; i < e[c][u].size() ; i++ ){int v = e[c][u][i];if ( vis[v] ) continue;vis[v] = true;q.push ( v );}}return false; }int main ( ) {while ( ~scanf ( "%d%d" , &n , &m ) ){for ( int i = 0 ; i < MAX ; i++ )for ( int j = 0 ; j < MAX ; j++ )e[i][j].clear();for ( int i = 0 ; i < m ; i++ ){scanf ( "%d%d%d" , &x , &y , &c );add ( x , y , c );}scanf ( "%d" , &q );while ( q-- ){int ans = 0;scanf ( "%d%d" , &x , &y );for ( int i = 1 ; i <= m ; i++ )if ( bfs ( i ) ) ans++;printf ( "%d\n" , ans );}} }

ac代碼3：（Floyd）

#include<bits/stdc++.h>using namespace std; const int MAX = 100 + 5;int maze[MAX][MAX][MAX]; bool bk[MAX]; int main() {int n,m;int a,b,c;int q,cnt;scanf("%d %d",&n,&m);while(m--) {scanf("%d %d %d",&a,&b,&c);maze[a][b][c]=1;maze[b][a][c]=1;bk[c]=1;}for(int c = 1; c<=MAX; c++) {//看看這個顏色有沒有出現過 //他確實是輸入m條路所以最多有m個顏色，但是你不能搜索到m啊！！ if(bk[c]==0) continue;for(int k = 1; k<=n; k++) {for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {if( maze[i][k][c]==1&&maze[k][j][c]==1) {maze[i][j][c]=1;}}}}}scanf("%d",&q);while(q--) {cnt=0;scanf("%d %d",&a,&b);for(int c = 1; c<=MAX; c++) {if(bk[c]==1 && maze[a][b][c] == 1) {cnt++;}}printf("%d\n",cnt);}return 0 ; }

總結：

?

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總結

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