Codeforces Round #286 (Div. 1) D. Mr. Kitayuta's Colorful Graph 并查集-优快云博客

本文介绍了一个涉及图论和数据结构的问题，即在一个包含多种颜色边的图中进行查询，找出两个顶点间直接或间接相连的颜色数量。文章通过并查集实现了一种高效的解决方案，并提供了完整的代码示例。

D. Mr. Kitayuta's Colorful Graph

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/506/problem/D

Description

Mr. Kitayuta has just bought an undirected graph with n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color c_i, connecting vertex a_i and b_i.

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

In the i-th query, he gives you two integers - u_i and v_i.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex u_i and vertex v_i directly or indirectly.

Input

The first line of the input contains space-separated two integers - n and m(2 ≤ n ≤ 10⁵, 1 ≤ m ≤ 10⁵), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers - a_i, b_i(1 ≤ a_i < b_i ≤ n) and c_i(1 ≤ c_i ≤ 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, (a_i, b_i, c_i) ≠ (a_j, b_j, c_j).

The next line contains a integer- q(1 ≤ q ≤ 10⁵), denoting the number of the queries.

Then follows q lines, containing space-separated two integers - u_i and v_i(1 ≤ u_i, v_i ≤ n). It is guaranteed that u_i ≠ v_i.

Output

For each query, print the answer in a separate line.

Sample Input

Sample Output

2
1
0

HINT

题意

给你一个图，然后有重边，每条边有一种颜色，然后每次查询两个点之间有多少种颜色是把这俩点直接相连的

题解：

每种颜色都直接用并查集来维护就好了

用map，就可以没必要初始化了

代码

#include<iostream>
#include<stdio.h>
#include<unordered_map>
using namespace std;
#define maxn 100007
unordered_map<int,int> H[maxn],ans[maxn];

int fi(int x,int y)
{
    return x==H[x][y]?x:H[x][y]=fi(H[x][y],y);
}
void uni(int x,int y,int c)
{
    if(H[x].find(c)==H[x].end())H[x][c]=x;
    if(H[y].find(c)==H[y].end())H[y][c]=y;
    int p = fi(x,c),q = fi(y,c);
    if(p==q)return;
    if(p>q)swap(p,q);
    H[fi(x,c)][c]=fi(y,c);
}

int main()
{
    int n,m;scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        if(x>y)swap(x,y);
        uni(x,y,z);
    }
    int q;scanf("%d",&q);
    while(q--)
    {
        int x,y;scanf("%d%d",&x,&y);
        if(H[x].size()>H[y].size())swap(x,y);
        if(ans[x].find(y)==ans[x].end())
        {
            int Ans = 0;
            for(auto &c:H[x])
            {
                if(H[y].find(c.first)==H[y].end())continue;
                int p = fi(x,c.first),q = fi(y,c.first);
                if(p==q)Ans++;
            }
            ans[x][y] = Ans;
        }
        printf("%d\n",ans[x][y]);
    }
}