ACM: Mr. Kitayuta's Colorful Graph-并查集-解题报

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

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.


题目还是比较水，很简单的并查集，思路清晰就够了；
AC代码：

+ View Code?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69

#include"iostream" #include"cstdio" #include"cstring" #include"cmath" #include"algorithm" using namespace std; const int MX=110; int CL[101]; struct nde {     int p[MX];  //记录每个颜色下的根 } cmap[MX];   struct node {     int a,b,c; //连接的点 a b 和颜色c } side[MX];   bool cmps(node a,node b) {     return a.c<b.c; }   struct nod {     int u,v;    //查询组 } Que[MX];   int find(int x,int c) {     return cmap[c].p[x]==x?x:(cmap[c].p[x]=find(cmap[c].p[x],c)); //找到在颜色C下的根。 }   void init() {               //清空并初始化整个颜色图。     for(int i=1; i<=100; i++)         for(int j=1; j<=100; j++) {             cmap[i].p[j]=j;         } }   int main() {     int n,m,q;     while(~scanf("%d%d",&n,&m)) {         init();         for(int i=1; i<=m; i++) {        //把整个输入先全部记录下             scanf("%d%d%d",&side[i].a,&side[i].b,&side[i].c);           }         sort(side+1,side+1+m,cmps);//按照颜色排序；         int tot=1;      //tot 记录颜色的种类         CL[tot]=side[1].c;//记录下出现过的颜色         for(int i=1; i<=m; i++) {             if(side[i].c!=CL[tot]) {    //如果下一个颜色与上一个颜色不同记录进数组  CL                 CL[++tot]=side[i].c;             }             int rt1=find(side[i].a,side[i].c);      //找到在颜色 C 下的根             int rt2=find(side[i].b,side[i].c);             if(rt1!=rt2) {                                  cmap[side[i].c].p[rt2]=rt1; //在颜色 C下将两点连接，更新根             }         }         scanf("%d",&q);         for(int i=1; i<=q; i++) {             scanf("%d%d",&Que[i].u,&Que[i].v);  //单组输入查询             int ans=0;             for(int j=1; j<=tot; j++) {          //每个出现过的颜色下都查询一次                 int rt1=find(Que[i].u,CL[j]);                   int rt2=find(Que[i].v,CL[j]);                 if(rt1==rt2)ans++;              //如果是一个联通块答案加一             }             printf("%d\n",ans);                 //输出该组的答案         }     }     return 0; }