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

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

Sample Output

102

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;}