Cube Stacking

DescriptionFarmer 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.Forcount 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.OutputPrint the output from each of the count operations in the same order as the input file.Sample Input

<span style="font-size:14px;">6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4
</span>

Sample Output

<span style="font-size:14px;">1
0
2
</span>

<span style="font-size:14px;"></span> 

<span style="font-size:14px;color:#006600;">题解：题意：n个方块（n<=30000）,p组操作（p<=100000）,操作有两种，M a b 将含有a的堆放在包含b的堆上，还有一种是 C a统计a下面有多少个方块,
思路：带权并查集，一堆中最顶上的方块作为父节点，用dist[X] 统计X到父亲节点的距离，rank[fa[X]]表示团的大小，两者相减即为答案。</span>

<span style="font-size:14px;"></span><pre class="cpp" name="code">#include<stdio.h>#include<string.h>int n;int per[32000];int dis[32000];//<span style="color:#009900;">dis[x]表示x到父节点的距离</span>int rank[32000];//<span style="color:#009900;">rank[x]表示以x为根结点的团的大小</span>void init(){for(int i=0;i<32000;i++){per[i]=i;dis[i]=0;rank[i]=1;}}int find(int x){if(x!=per[x]){int t=per[x];per[x]=find(per[x]);dis[x]+=dis[t];}return per[x];}//<span style="color:#009900;">在找x根节点的同时，也计算出了x到根节点的距离。</span>int main(){char ch;while(scanf("%d",&n)!=EOF){init();while(n--){scanf("%c",&ch);getchar();int x,y,a,b;if(ch=='M'){scanf("%d%d",&a,&b);int fx=find(a);int fy=find(b);if(fx!=fy){per[fy]=fx;//<span style="color:#009900;">让a的根节点成为b的根节点，实现了含有a的堆放在含有b的堆上面。</span>dis[fy]=rank[fx];//<span style="color:#009900;">计算b的根节点到新根节点的距离</span>		    		rank[fx]+=rank[fy];//<span style="color:#009900;">计算以fx为根节点的团的大小</span>		    	}}else{scanf("%d",&a);int fa=find(a);printf("%d\n",rank[fa]-dis[a]-1);//<span style="color:#009900;">a的根节点所在的团的大小 减去 a到根节点的个数 减去 1。</span>}}}return 0;}