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hdu 2818 Building Block

2024-08-08 14:40:14   221人阅读

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.
 

 

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
6M 1 6C 1M 2 4M 2 6C 3C 4
 

 

Sample Output
102
 

 

Source
2009 Multi-University Training Contest 1 - Host by TJU

 

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\n",under[a]);61       }62 63      }64 65  }66  return 0;67 }

 

hdu 2818 Building Block


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