CodeForces 505B Mr. Kitayuta's Colorful Graph
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 — 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.
Sample 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
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
Hint
Let's consider 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.
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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,0,sizeof(vis)); while(!q.empty()) { now=q.front(); if(now==y) return true; q.pop(); for(int i=0;i<edg[c][now].size();i++) { int then=edg[c][now][i]; if(vis[then]) continue; vis[then]=1; q.push(then); } } return false; } int main() { int n,m; scanf("%d%d",&n,&m); int a,b,c; for(int i=0;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=0;i<q;i++) { ans=0; scanf("%d%d",&x,&y); for(int i=0;i<m;i++) if(bfs(i+1)) 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。