CodeForces-505B-Mr. Kitayuta's Colorful Graph（弗洛伊德）

H - Mr. Kitayuta’s Colorful Graph
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status

Practice

CodeForces 505B
Description
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 Input
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
Hint
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：首行给出N和M，代表有N个点。接下来有M行，每行三个数字A,B,C，代表A到B有颜色为C的绳（也可以是线，随便是什么吧）。接着给出Q，代表Q次询问。接下来Q行，每行给出A和B，只有相同颜色的绳才可以把AB联通，问联通AB的绳子一共有多少种。

思路：map[p][i][j]=1代表i到j被颜色p联通

代码（可以优化的地方很多，不过后台水，就这么过去了）

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>using namespace std;const int maxn=101;int map[maxn][maxn][maxn];//初始化为0int num[maxn];//辅助记录i是否出现过int N;//N个点int M;//M条边int Q;//Q次询问int main(){    memset(map,0,sizeof(map));    memset(num,0,sizeof(num));    scanf("%d%d",&N,&M);    int a,b,c;    while(M--)//建图    {        scanf("%d%d%d",&a,&b,&c);        map[c][a][b]=1;//a到b以颜色c联通        map[c][b][a]=1;        //是否记录颜色最值，视情况优化        num[c]=1;//标记此颜色出现过    }    for(int p=1; p<=maxn; p++)//弗洛伊德更新连通性    {        if(num[p])//颜色p出现过        {            for(int k=1; k<=N; k++)            {                for(int i=1; i<=N; i++)                {                    for(int j=1; j<=N; j++)                    {                        if(!map[p][i][j]&&map[p][i][k]&&map[p][k][j])                        {                            map[p][i][j]=1;                        }                    }                }            }        }    }    scanf("%d",&Q);//    //打印1到4//    for(int p=1;p<=3;p++)//    {//        for(int i=1;i<=4;i++)//        {//            for(int j=1;j<=4;j++)//            {//                printf("%d ",map[p][i][j]);//            }//            printf("\n");//        }//        printf("\n\n");//    }    while(Q--)    {        int count=0;        scanf("%d%d",&a,&b);        for(int p=1; p<=maxn; p++)        {            if(num[p]==1&&map[p][a][b]==1)            {                count++;            }        }        printf("%d\n",count);    }    return 0;}