Codeforces Round #286 (Div. 2) B. Mr. Kitayuta's Colorful Graph +foyd算法的应用

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.

解决方案：题目求的是先建立一个图，有很多种不同颜色的边，每次询问两条边的不同颜色的路径的个数。由此联想的floyd算法，可以枚举边的颜色，求每种颜色的最短路，然后在统计不同颜色的最短路的个数。

code：

<pre name="code" class="cpp">#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int Mat[102][102][102];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(Mat,0,sizeof(Mat));
        for(int i=1; i<=m; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            Mat[a][b][c]=1;
            Mat[b][a][c]=1;
        }
        for(int c=1;c<=m;c++){
            for(int k=1;k<=n;k++){
                for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++){
                    if(Mat[i][k][c]&&Mat[k][j][c]){
                        Mat[i][j][c]=1;
                        Mat[j][i][c]=1;
                    }
                }
            }
        }
        int q;
        scanf("%d",&q);
        for(int i=0;i<q;i++){
            int a,b;
            int sum=0;
            scanf("%d%d",&a,&b);
            for(int i=1;i<=m;i++){
                sum+=Mat[a][b][i];
            }
            printf("%d\n",sum);
        }
    }
    return 0;
}

Codeforces Round #286 (Div. 2) B. Mr. Kitayuta's Colorful Graph +foyd算法的应用