hdu 2818 Building Block

Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5811    Accepted Submission(s): 1790

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

 

用一个数组记录每块积木上面的积木数，再用一个数组记录每块积木下面的积木数，将每堆积木的最下面一块积木作为上面每块积木的根节点，每次操作的时候去更新这三个数组数值就好了。

#include<iostream>#include<cstring>#include<cmath>#include<algorithm>#include<cstdio> using namespace std;int city,road;int pre[33333];int up[33333],down[33333];int find(int x){if(x==pre[x])return x;    int tem=find(pre[x]);    down[x]=down[x]+down[pre[x]];    return pre[x]=tem;/*int i=x,j;   while(pre[i]!=r)      {          j=pre[i];          pre[i]=r;          i=j;      } */}void merge(int x,int y){    int fx=find(x);    int fy=find(y);    if(fx!=fy)    pre[fy]=fx;}int main(){    int n;    while(cin>>n)    {        int i;        char op[11];        int a,b;        for(i=0;i<=31015;i++)        {            pre[i]=i;            up[i]=1;            down[i]=0;        }        for(i=1;i<=n;i++)        {            cin>>op;            if(op[0]=='M')            {                scanf("%d%d",&a,&b);                int fx=find(a);                int fy=find(b);                if(fx!=fy)                {                    pre[fx]=fy;                    down[fx]=up[fy];                    up[fy]+=up[fx];                }            }            else            {                scanf("%d",&a);                find(a);                cout<<down[a]<<endl;            }        }    }    return 0;}