Cube Stacking（POJ-1988）

Description
Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There
are two types of operations:

moves and counts.

* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.

* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.

Write a program that can verify the results of the game.

Input
* Line 1: A single integer, P

* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For
count operations, the line also contains a single integer: X.

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.

Output
Print the output from each of the count operations in the same order as the input file.

Sample Input

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

Sample Output

1
0
2

题意：搬箱子和架箱子，如果箱子上面有箱子，则再不打乱顺序的条件下把整体搬过去

思路：在并查集的基础上加一个旧根的深度的数组，旧根的深度等于新根上一次的节点数，最后输出的时候，查根，减去改节点的深度，再减去本身

#include<iostream>
#include<cstdio>
using namespace std;
int a[30005];
int d[30005];
int f[30005];
int find(int x)
{
int temp;
if(x == a[x])
return x;
temp = a[x];
a[x] = find(temp);
d[x] += d[temp];
return a[x];
}
int main()
{
int p;
int x,y;
char s;
char c;
scanf("%d",&p);
s=getchar();
for(int i=0;i<=30000;i++){
a[i]=i;
d[i]=0;
f[i]=1;
}
for(int i=0;i<p;i++){
scanf("%c",&c);
if(c=='M'){
scanf("%d%d",&x,&y);
s=getchar();
x=find(x);
y=find(y);
if(x!=y){
a[y]=x;
d[y]=f[x];
f[x]+=f[y];
}
}
if(c=='C'){
scanf("%d",&x);
s=getchar();
printf("%d\n",f[find(x)]-d[x]-1);
}
}
return 0;
}