Cube Stacking 1988 （并查集 好题）

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 21402		Accepted: 7492
Case Time Limit: 1000MS

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

//好题：：运用并查集
/*题意：：
有N个立方体（筛子），从1到N编号，现在用这些立方体（筛子）做一个游戏
输入一个t,表示有几行，再输入一个字符，若字符是‘M’，则输入两个数，表示两个筛子的编号x,y，表示将
x放到y上面 ，若字符为‘C’，则输入一个数，表筛子编号，然后输出这个筛子下面有几个筛子。*/

//例如：：输入样例（这块看了半天没看懂）
/*   M  1  6  表示将1放到6上面

1
6

C   1   表示输出1下面有几个数，很容易看出有一个
M   2  4  表示将2放到4上

2
4

M   2  6  ***这块重点，表示将2放到6上

2     他每次输入都是连续的，即使有输出也不会中断，直到输入的行数等于 t 。
4
1
6

C   4  表示4下面有几个数，可以看出有2个。*/
//集体解题思路见代码
<pre class="cpp" name="code">#include<stdio.h>
#include<string.h>
#include<math.h>
int a[30110];
int b[30110];//记录从该节点到最后节点的深度
int s[30110];//记录节点所处的深度
int find(int x)
{
if(x!=a[x])
{
int t=a[x];
a[x]=find(a[x]);
s[x]+=s[t];
}
return a[x];
}
int marge(int x,int y)
{
int fx,fy;
fx=find(x);
fy=find(y);
if(fx!=fy)
{
a[fy]=fx;
s[fy]=b[fx];//这块是重点，自己多找几个点模拟一下就清楚了
b[fx]+=b[fy];
}
}
int main(){
int t;
while(scanf("%d",&t)!=EOF)
{
int i,x,y;
char c;
for(i=0;i<=30000;i++)
{
a[i]=i;
b[i]=1;
s[i]=0;
}
while(t--)
{
getchar();
scanf("%c",&c);
if(c=='M')
{
scanf("%d%d",&x,&y);
marge(x,y);
}
else
{
scanf("%d",&x);
int dx=find(x);//找到根节点
printf("%d\n",b[dx]-s[x]-1);//从根节点到要找的节点间的节点数
}
}
}
return 0;
}