B. Mr. Kitayuta's Colorful Graph（ Codeforces Round #286 (Div. 2)）

B. Mr. Kitayuta's Colorful Graph

time limit per test

memory limit per test

input

output

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.

q

i-th query, he gives you two integers — ui and vi.

ui and vertex vi

Input

n and m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100), denoting the number of the vertices and the number of the edges, respectively.

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).

q (1 ≤ q ≤ 100), denoting the number of the queries.

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.

• 1 and vertex 2 are connected by color 1 and 2.
• 3 and vertex 4 are connected by color 3.
• 1 and vertex 4

并查集做法：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int bin[210][210];
int n,m;

void init()
{
    for(int i=1;i<201;i++)
    {
        for(int j=1;j<201;j++)
        {
            bin[i][j] = j;
        }
    }
}

int findx(int z,int x)
{
    while(bin[z][x]!=x)
    {
        x = bin[z][x];
    }
    int k = x,j;
    while(k!=x)
    {
        j = bin[z][k];
        bin[z][k] = x;
        k = j;
    }
    return x;
}

void merrg(int z,int x,int y)
{
    int fx = findx(z,x);
    int fy = findx(z,y);
    if(fx!=fy)
    {
        bin[z][fx] = fy;
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        int x,y,z;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            merrg(z,x,y);
        }
        int k;
        int a,b;
        scanf("%d",&k);
        while(k--)
        {
            int cnt = 0;
            scanf("%d%d",&a,&b);
            for(int i=1;i<=m;i++)
            {
                int aa = findx(i,a);
                int bb = findx(i,b);
                if(aa == bb)
                {
                    cnt++;
                }
            }
            printf("%d\n",cnt);
        }
    }
    return 0;
}

floyd做法：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int map[101][101][101];
int n,m;

void init()
{
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=n;j++)
        {
            for(int k=1;k<=n;k++)
            {
                map[i][j][k] = 0;
            }
        }
    }
}

void flordy()
{
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=n;j++)
        {
            for(int k=1;k<=n;k++)
            {
                for(int p=1;p<=n;p++)
                {
                    //printf("map[%d][%d][%d] = %d    map[%d][%d][%d] = %d\n",i,k,j,map[i][k][j],i,j,p,map[i][j][p]);
                    if(map[i][k][j] == 1 && map[i][j][p] == 1)
                    {
                        map[i][k][p] = 1;
                        map[i][p][k] = 1;
                    }
                }
            }
        }
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        int x,y,z;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            map[z][x][y] = 1;
            map[z][y][x] = 1;
        }
        flordy();
        int t;
        int a,b;
        scanf("%d",&t);
        while(t--)
        {
            int cnt = 0;
            scanf("%d%d",&a,&b);
            for(int i=1;i<=m;i++)
            {
                if(map[i][a][b] == 1)
                {
                    cnt++;
                }
            }
            printf("%d\n",cnt);
        }
    }
    return 0;
}