Codeforces Round #286 (Div. 2) B. Mr. Kitayuta's Colorful Graph dfs
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.
![](http://codeforces.com/predownloaded/bc/ea/bcea0be251489a112c0eff5bfb76b4310a7aa5a2.png)
- 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.
题意:给你一个图n个点,m条边;
每条边有个颜色,每次u,v只能走一条颜色的路径;
求有多少条路;
思路:暴力dfs;
#include<bits/stdc++.h> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-14http://v.ifeng.com/program/lyyy/ const int N=2e5+10,M=1e6+10,inf=1e9+10,MOD=2009; const ll INF=1e18+10; struct edge { int v,w; int nex; }edge[N]; int head[N],edg; void init() { memset(head,-1,sizeof(head)); edg=0; } void add(int u,int v,int w) { ++edg; edge[edg].v=v; edge[edg].w=w; edge[edg].nex=head[u]; head[u]=edg; } int now,to,ans; int flag[N]; int hh[N]; void dfs(int u,int pre) { //cout<<u<<" "<<pre<<endl; if(u==to) { if(!hh[pre]&&pre!=-1) { ans++; hh[pre]=1; } return; } for(int i=head[u];i!=-1;i=edge[i].nex) { int v=edge[i].v; if(flag[v])continue; int w=edge[i].w; if(pre==-1||w==pre) { flag[v]=1; dfs(v,w); flag[v]=0; } } } int main() { init(); int n,m; scanf("%d%d",&n,&m); for(int i=0;i<m;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); add(u,v,w); add(v,u,w); } int q; scanf("%d",&q); while(q--) { memset(hh,0,sizeof(hh)); ans=0; scanf("%d%d",&now,&to); flag[now]=1; dfs(now,-1); flag[now]=0; printf("%d\n",ans); } return 0; }