Building Block
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1982 Accepted Submission(s): 598
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
Recommend
gaojie
题目大意:有n个blocks,定义两种操作，一种是M x y 将block x加入到block y 下面，包括x的下面所有blocks。另一个是C x，计算block x 下面block 的数量并输出来。
分析：这道题可以通过并查集来解决，首先每个block的跟自己归在一类，，当遇到M操作时，将x 集合加入到y集合下面，并将y的数量加上x的数量，当碰到C 操作时，只要输出相应的数量即可。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #define maxlen 30010 5 using namespace std; 6 int father[maxlen],all[maxlen],sum[maxlen]; 7 void Init() 8 { 9 for(int i=0; i<maxlen; ++i) 10 { 11 father[i]=i; 12 all[i]=1; 13 sum[i]=0; 14 } 15 } 16 17 int Find(int x) 18 { 19 if(x == father[x]) 20 return father[x]; 21 int temp=Find(father[x]); 22 sum[x]+=sum[father[x]]; 23 father[x]=temp; 24 return father[x]; 25 } 26 27 void Union(int x,int y) 28 { 29 int a=Find(x); 30 int b=Find(y); 31 if(a==b) 32 return; 33 father[a]=b; 34 sum[a]+=all[b]; 35 all[b]+=all[a]; 36 } 37 38 int main() 39 { 40 int n,a,b; 41 char s; 42 Init(); 43 scanf("%d",&n); 44 while(n--) 45 { 46 getchar(); 47 scanf("%c",&s); 48 if(s == 'M') 49 { 50 scanf("%d%d",&a,&b); 51 Union(a,b); 52 } 53 else 54 { 55 scanf("%d",&a); 56 Find(a); 57 printf("%d ",sum[a]); 58 } 59 } 60 return 0; 61 }