CodeForces - 505B Mr. Kitayuta's Colorful Graph（二维并查集）

最新推荐文章于 2024-01-28 20:34:04 发布

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最新推荐文章于 2024-01-28 20:34:04 发布

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分类专栏：并查集

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博客讨论了如何使用二维并查集解决CodeForces上的505B问题，即在一个无向图中处理颜色相关的查询。题目要求找出两个给定点之间通过直接或间接的特定颜色边连接的数量。输入包括图的顶点数、边数、每条边的颜色和查询次数，输出是每次查询的答案。由于数据规模较小，可以采用并查集的数据结构来有效处理各个颜色的连通性。

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题目链接：http://codeforces.com/problemset/problem/505/B点击打开链接

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.

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.

数据量很小可以用并查集主要就是构造每个颜色的并查集这样才能区别每个地点是否与上一个地点相连

#include <bits/stdc++.h>
using namespace std;
#define maxn 200
int pre[maxn][maxn];
int findx(int col,int x)
{
    int r=x;
    while(pre[col][r]!=r)
    {
        r=pre[col][r];
    }
    int i=x;int j;
    while(pre[col][i]!=r)
    {
        j=pre[col][i];
        pre[col][i]=r;
        i=j;
    }
    return r;
}

void join (int col,int x,int y)
{
    int p1=findx(col,x);
    int p2=findx(col,y);
    if(p1!=p2)
    {
        pre[col][p2]=p1;
    }
}
int judge(int col,int x,int y)
{
    if(findx(col,x)!=findx(col,y))
        return 0;
    else
        return 1;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i<maxn;i++)
        for(int j=0;j<maxn;j++)
            pre[i][j]=j;
    for(int i=0;i<m;i++)
    {
        int mid1,mid2,mid3;
        scanf("%d%d%d",&mid1,&mid2,&mid3);
        join(mid3,mid1,mid2);
    }
    int t=0;
    scanf("%d",&t);
    for(int i=0;i<t;i++)
    {
        int mid1,mid2;
        scanf("%d%d",&mid1,&mid2);
        int ans=0;
        for(int i=1;i<=m;i++)
            ans+=judge(i,mid1,mid2);
        printf("%d\n",ans);
    }
}