HDU2818 并查集

Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5341    Accepted Submission(s): 1638

Problem Description

John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command. 
C X : Count the number of blocks under block X

You are request to find out the output for each C operation.

 

Input

The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

 

Output

Output the count for each C operations in one line.

 

Sample Input

6

M 1 6

C 1

M 2 4

M 2 6

C 3

C 4

 

Sample Output

1

0

2

 Source

2009 Multi-University Training Contest 1 - Host by TJU

题意：

p次操作，M x y 表示将x所在的一堆所有数放到y所在的一堆上面，C x 表示询问数字x下方有几个数字。

代码：

 #include<iostream>
 #include<cstdio>
 using namespace std;
 int fat[],num[],sum[];//num[i]记录数i下面有几个数，sum[i]记录以i为根节点的并查集有几个元素
 void init()
 {
     for(int i=;i<=;i++){
         fat[i]=i;
         num[i]=;
         sum[i]=;
     }
 }
 int find(int x)
 {
     if(fat[x]!=x){
         int tem=fat[x];
         fat[x]=find(tem);
         num[x]+=num[tem];
     }
     return fat[x];
 }
 void connect(int x,int y)
 {
     int a=find(x),b=find(y);
     if(a!=b){
         fat[a]=b;
         num[a]=sum[b];
         sum[b]+=sum[a];
     }
 }
 int main()
 {
     int p,x,y;
     char ch;
     while(scanf("%d",&p)!=EOF){
         init();
         while(p--){
             cin>>ch;
             if(ch=='M'){
                 scanf("%d%d",&x,&y);
                 connect(x,y);
             }
             else{
                 scanf("%d",&x);
                 find(x);  //更新一下num
                 printf("%d\n",num[x]);
             }
         }
     }
     return ;
 }