【杭电oj】3635 - Dragon Balls（带权并查集，好题）

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5230    Accepted Submission(s): 1964

Problem Description

Five hundred years later, the number of dragon balls will increase unexpectedly, so it's too difficult for Monkey King(WuKong) to gather all of the dragon balls together. 

His country has N cities and there are exactly N dragon balls in the world. At first, for the ith dragon ball, the sacred dragon will puts it in the ith city. Through long years, some cities' dragon ball(s) would be transported to other cities. To save physical strength WuKong plans to take Flying Nimbus Cloud, a magical flying cloud to gather dragon balls. 
Every time WuKong will collect the information of one dragon ball, he will ask you the information of that ball. You must tell him which city the ball is located and how many dragon balls are there in that city, you also need to tell him how many times the ball has been transported so far.

 

Input

The first line of the input is a single positive integer T(0 < T <= 100). 
For each case, the first line contains two integers: N and Q (2 < N <= 10000 , 2 < Q <= 10000).
Each of the following Q lines contains either a fact or a question as the follow format:
  T A B : All the dragon balls which are in the same city with A have been transported to the city the Bth ball in. You can assume that the two cities are different.
  Q A : WuKong want to know X (the id of the city Ath ball is in), Y (the count of balls in Xth city) and Z (the tranporting times of the Ath ball). (1 <= A, B <= N)

 

Output

For each test case, output the test case number formated as sample output. Then for each query, output a line with three integers X Y Z saparated by a blank space.

 

Sample Input

23 3T 1 2T 3 2Q 23 4T 1 2Q 1T 1 3Q 1

 

Sample Output

Case 1:2 3 0Case 2:2 2 13 3 2

 

Author

possessor WC

 

Source

2010 ACM-ICPC Multi-University Training Contest（19）——Host by HDU

还是带权并查集的题，这道题还记录了节点的个数，其实还是跟带权并查集的题一样。

这道题注意初始化数据。

代码如下：

#include <cstdio>#include <algorithm>using namespace std;int f[10022],m[10022];//表示压缩路径的次数（转移次数） int num[10022];//节点数 int n,q;void init(){for (int i = 1 ; i <= n ; i++){f[i] = i;num[i] = 1;//自己也算m[i] = 0; }}int find (int x){if (x != f[x]){int t = f[x];f[x] = find (f[x]);m[x] += m[t];//仍然可以看做带权并查集 ，只是这个数据需要在其他地方处理 }return f[x];}void join(int x,int y){int fx,fy;fx = find (x);fy = find (y);if (fx != fy){f[fx] = fy;//根节点连过去num[fy] += num[fx];//把节点数也算过去m[fx] = 1;//记录节点移动了（子节点后期在路径压缩中会加上该数值） }return;}int main(){int u;int k = 1;char op[2];scanf ("%d",&u);while (u--){scanf ("%d %d",&n,&q);printf ("Case %d:\n",k++);init();//初始化别忘了 for (int i = 1 ; i <= q ; i++){scanf ("%s",op);if (op[0] == 'T'){int x,y;scanf ("%d %d",&x,&y);join(x,y);}else{int t;scanf ("%d",&t);find(t);//保证最后一次也压缩了路径，这里就起压缩作用 printf("%d %d %d\n",find(t),num[find(t)],m[t]);}}}return 0;}