题干：
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.
解题报告：
ac代码1：（并查集）（15ms，0.1MB）
原来的代码加上了ra数组，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代码2：（广搜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代码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++) {//看看这个颜色有没有出现过 //他确实是输入m条路所以最多有m个颜色，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 ;
}总结：
