B. Mr. Kitayuta's Colorful Graph (CF #286 (Div. 2) 并查集)

B. Mr. Kitayuta‘s Colorful Graph

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

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

output

2
1
0

input

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

output

1
1
1
1
2

Note

Let‘s consider the first sample.

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：现有n个点m条边的无向图，每条边都有一种颜色，然后有q次询问，x y询问点x到点y共有几种颜色的边将他们连起来（不同颜色的边不能混在一起，要分开看）

思路：二维并查集，每种颜色维护一个并查集，查询时看某一种颜色下两个点是否有共同的father。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
typedef long long ll;
using namespace std;

int n,m;
int color[110];
int father[110][110];

void init()
{
    for (int i=0;i<=110;i++)
        for (int j=0;j<=110;j++)
            father[i][j]=j;
}

int find_father(int c,int x)        //递归写法
{
    if (x!=father[c][x])
        father[c][x]=find_father(c,father[c][x]);
    return father[c][x];
}

//int find_father(int c,int x)  //非递归写法
//{
//    int r=x,i,j;
//    while(r!=father[c][r])
//        r=father[c][r];
//    i=x;
//    while(i!=r)
//    {
//        j=father[c][i];
//        father[c][i]=r;
//        i=j;
//    }
//    return r;
//}

int main()
{
    while (~scanf("%d%d",&n,&m))
    {
        init();
        int a,b,c;
        memset(color,0,sizeof(color));  //记录下用了哪些颜色
        for (int i=0;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            color[c]=1;
//            printf("+++\n");
            int fa=find_father(c,a);
            int fb=find_father(c,b);
            if (fa!=fb)
                father[c][fa]=fb;
        }
        int q,x,y;
        scanf("%d",&q);
        while (q--)
        {
            scanf("%d%d",&x,&y);
            int ans=0;
            for (int i=1;i<=m;i++)
            {
                if (color[i])
                {
                    if (find_father(i,x)==find_father(i,y))
                        ans++;
                }
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}

B. Mr. Kitayuta's Colorful Graph (CF #286 (Div. 2) 并查集)