Cube Stacking
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
题目大意：有两种操作：M x y 把包含x的栈放到含y的栈上 C x 查询含x栈中x下面有几个数
思路：并查集合并栈，用两个数组记录栈中所有元素个和在x上面的个数
x所在栈中所有元素个数 - 在x上面的个数 - x本身 就是x下面的元素个数
代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N=100000+20;
int fa[N],d[N],c[N]; //c记录所有元素个数，d记录在它上面的元素个数
int find(int x)
{
int fx=fa[x];
if(fa[x]==x) return x;
fx=find(fa[x]);
d[x]+=d[fa[x]];
return fa[x]=fx;
}
void join(int x,int y)
{
int fx,fy;
fx=find(x),fy=find(y);
if(fx!=fy)
fa[fy]=fx;
d[fy]+=c[fx];
c[fx]+=c[fy];
}
int main()
{
int n;
while(~scanf("%d%d",&n))
{
char s;
int x,y;
for(int i=1;i<=n;i++){fa[i]=i;c[i]=1;d[i]=0;}
for(int i=0;i<n;i++)
{
scanf(" %c",&s);
if(s=='M')
{
scanf("%d%d",&x,&y);
join(x,y);
}
else
{
scanf("%d",&x);
y=find(x);
printf("%d\n",c[y]-d[x]-1);
}
}
}
return 0;
}