Building Block
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3250 Accepted Submission(s): 973
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.
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
题意：
1 /*有N个砖头， 编号1—N(实际上是0-N)，然后有两种操作，第一种是M x y 把x所在的那一堆砖头全部移动放到y所在的那堆上面。 第二种操作是 C x ，即查询x下面有多少个砖头 ，并且输出。 2 */
--->带权值的并查集
代码：
1 #include<cstring> 2 #include<cstdio> 3 #include<cstdlib> 4 #define maxn 30030 5 using namespace std; 6 int father[maxn]; 7 __int64 rank[maxn],under[maxn]; 8 int p; 9 10 void init(){ 11 12 for(int i=0;i<maxn ;i++) { 13 father[i]=i; 14 rank[i]=1; 15 under[i]=0; 16 } 17 18 } 19 20 int fin(int x){ 21 22 if(x == father[x]) 23 return father[x]; 24 int tem = father[x] ; 25 father[x] = fin(father[x]); 26 under[x]+=under[tem]; 27 28 return father[x]; 29 30 } 31 32 void Union(int a,int b){ 33 34 int x = fin(a); 35 int y = fin(b); 36 if( x==y ) return ; 37 father[x] = y ; //将a所在的堆放在b的堆上 38 under[x] = rank[y]; 39 rank[y] += rank[x]; 40 rank[x] = 0; 41 42 } 43 44 int main() 45 { 46 char str[2]; 47 int a,b; 48 49 while(scanf("%d",&p)!=EOF){ 50 init(); 51 while(p--){ 52 scanf("%s",str); 53 if(str[0]=='M'){ 54 scanf("%d%d",&a,&b); 55 Union(a,b); 56 } 57 else{ 58 scanf("%d",&a); 59 fin(a); 60 printf("%I64d ",under[a]); 61 } 62 63 } 64 65 } 66 return 0; 67 }