Dragon Balls--hdu3635（并查集）

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4346    Accepted Submission(s):
1658

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

2

3 3

T 1 2

T 3 2

Q 2

3 4

T 1 2

Q 1

T 1 3

Q 1

Sample Output

Case 1:

2 3 0

Case 2:

2 2 1

3 3 2

这个题大意是开始的时候一个城市一个龙珠，如果输入T A B就是把A球所在的城市中的球全部转移到B球所在的城市

输入Q A就是要输出A球所在的城市序号，A球所在城市的球的个数，A球移动的次数

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define max 10002
 using namespace std;
 int father[max],ran[max],num[max];
 int a,b;

 void init()
 {
     int i;
     for(i=;i<=a;i++)
     {
         father[i]=i;//记录父节点
         ran[i]=;//记录所在城市共有多少龙珠
         num[i]=;//记录移动的次数
     }
 }

 int find(int x)
 {
     if(x==father[x]) return x;
     int t=father[x];
     father[x]=find(father[x]);//压缩路径 ，都指向根节点
     num[x]+=num[t];//每个球移动的次数等于本身移动的个数加上父节点移动的次数
     return father[x];
 }

 void join(int x,int y)
 {
     int fx=find(x);
     int fy=find(y);
     if(fx!=fy)
     {
         father[fx]=fy;//移动一个根节点到另一个根节点
         ran[fy]+=ran[fx];//fy为根节点的总个数等于两个根节点拥有的个数相加
         num[fx]=;//被移动的根节点第一次移动
     }
 }

 int main()
 {
     int i,N,m,n,cot=;
     char c;
     scanf("%d",&N);
     while(N--)
     {
         scanf("%d%d",&a,&b);
         getchar();
         init();
         printf("Case %d:\n",cot++);
         for(i=;i<b;i++)
         {
             scanf("%c",&c);
             if(c=='T')
             {
                 scanf("%d%d",&m,&n);
                 getchar();
                 join(m,n);
             }
             else
             {
                 scanf("%d",&m);
                 getchar();
                 int t=find(m);//要输出根节点的所有的个数
                 printf("%d %d %d\n",t,ran[t],num[m]);
             }
         }
     }
     return ;
 }