CodeForces - 505B 图的连同性，多维并查集或用搜索写-优快云博客

本文介绍了一种使用并查集解决图论中涉及颜色路径连接的问题，包括两种方法：搜索bfs和多维并查集。通过具体实例展示了如何实现这两种算法。

Mr. Kitayuta has just bought an undirected graph consisting of n vertices and medges. 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 ≤ 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 — 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 ≤ 100), 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.

Example

Input

Output

2
1
0

Input

Output

Note

Let's consider the first sample.

$\text{[math]}$ 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.

题意：

输入你q个问题，下面q行，每行有两个点，找取两个点有几条路连通，每种颜色代表一条路；

小编因为英语的硬伤，没有理解清题意，以为两个颜色相同就行了，谁知道还要判断连通性；

在这里小编提醒一下各位博友，做题时一定要审清题；

方法一：搜索 bfs

#include<iostream>   //知道前面为啥一直错吗，题意不理解 ,颜色相同，还得必须连通； 
#include<stdio.h>		// 这是用搜索做的； 
#include<string.h>		// （1）以后做题先申清题 ，审不清题，就别做； 
#include<vector>		
#include<queue>	
#include<algorithm>
using namespace std;

struct node 
{
	int e,num;
};

vector<node> v[110];

int n,m,sum,x,y,book[110][110];

void init()
{
	for(int i=0;i<=n;i++)
		v[i].clear();
}

void bfs()
{
	 queue<node> q;
	 int i;
	 for(i=0;i<v[x].size();i++)
	 {
	 	node st=v[x][i];
	 	book[st.e][st.num]=1;
	 	q.push(st);
	 }
	 while(!q.empty())
	 {
	 	node st=q.front();
	 	q.pop();
	 	if(st.e==y)
	 	{
	 		sum++;
	 		continue;
		}
		for(i=0;i<v[st.e].size();i++)
		{
			node end=v[st.e][i];
			if(!book[end.e][end.num]&&st.num==end.num)
			{
				book[end.e][end.num]=1;
				q.push(end);
			}
		}
	 }
}
int main()
{
	int i,j,k,t;
	while(~scanf("%d%d",&n,&m))
	{
		init();
		int x1,y1,v1;
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&x1,&y1,&v1);
			node st;
			st.e=y1;
			st.num=v1;
			v[x1].push_back(st);
			st.e=x1;
			v[y1].push_back(st);
		}
		scanf("%d",&t);
		while(t--)
		{
			sum=0;
			memset(book,0,sizeof(book));
			scanf("%d%d",&x,&y);
			fill(book[x],book[x]+101,1);
			bfs();
			printf("%d\n",sum);
		}
	}
	return 0;
}

方法二：

多维并查集

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m,f[110][110];
void init()
{
	for(int i=0;i<=m;i++)
		for(int j=0;j<=n;j++)
			f[i][j]=j;
}

int find(int x,int v)
{
	if(f[v][x]==x)
		return x;
	else 
	{
		f[v][x]=find(f[v][x],v);
		return f[v][x];
	}
}

int main()
{
	int i,j,k,t;
	while(~scanf("%d%d",&n,&m))
	{
		init();
		int x1,y1,v1;
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&x1,&y1,&v1);
			int t1=find(x1,v1);
			int t2=find(y1,v1);
			if(t1<t2)
				f[v1][t2]=t1;
			else f[v1][t1]=t2;
		}
		int x,y;
		scanf("%d",&t);
		for(i=0;i<t;i++)
		{
			scanf("%d%d",&x,&y);
			int sum=0;
			for(j=1;j<=m;j++)
			{
				if(find(x,j)==find(y,j))
					sum++;
			}
			printf("%d\n",sum);
		}
	}
	return 0;
}