Codeforces 505B __Mr. Kitayuta's Colorful Graph(floyd)

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 51 2 11 2 22 3 12 3 32 4 331 23 41 4

output

210

input

5 71 5 12 5 13 5 14 5 11 2 22 3 23 4 251 55 12 51 51 4

output

11112

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.

题意看了hint就很明了了。

解法： 把点a，b，c用一个三维数组来表示。存在为1，不存在为0。接着用floyd来把那些间接通过一个颜色相连的点连在一起，最后查询即可。

#include<cstdio>#include<cstring>int vis[110][110][110];int main(){    int n,m,a,b,c;    while(~scanf("%d%d",&n,&m)) {        memset(vis,0,sizeof(vis));        for(int i=0;i<m;i++) {            scanf("%d%d%d",&a,&b,&c);            vis[a][b][c]=1;            vis[b][a][c]=1;        }        for(int i=1;i<=m;i++) {            for(int k=1;k<=n;k++) {                for(int a=1;a<=n;a++) {                    for(int b=1;b<=n;b++) {                        if(vis[a][k][i]&&vis[k][b][i]) {                            vis[a][b][i]=vis[b][a][i]=1;                        }                    }                }            }        }        int q;        scanf("%d",&q);        for(int i=0;i<q;i++) {            scanf("%d%d",&a,&b);            int ans=0;            for(int j=1;j<=m;j++) {                if(vis[a][b][j]) {                    ans++;                }            }            printf("%d\n",ans);        }    }    return 0;}