POJ

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

10
2
题意：给你1~n个木块初始状态是n堆，每堆一个，t次操作，如果是M的话，就是把a这一堆放到b堆的上面，如果是C的话，就是查询a下面有多少个木块。
思路，并查集，不过要记录某一堆的总木块数量，sum[],方便合并的时候，计算某个木块下面的木块数量，还要记录每个木块下面的数量，数组numunder[]
#include<iostream>#include<cstdio>using namespace std;#define N 30005int father[N];int sum[N];int numunder[N];void init(){    int i;    for(i=0;i<=N;i++){        father[i]=i;        sum[i]=1;        numunder[i]=0;    }}int Find(int x){    if(x!=father[x]){        int f=father[x];        father[x]=Find(father[x]);        numunder[x]+=numunder[f];//递归去更新每个的下面木块的数量    }    return father[x];}void Move(int a,int b){    int f1=Find(a);    int f2=Find(b);    father[f1]=f2;//是将其下面的木块当成父亲    numunder[f1]=sum[f2];//f1堆下面的木块量是f2堆的总量    sum[f2]+=sum[f1];//合并之后f2堆的总数变化}int main(){    init();//初始化    int t;    char ch;    int a,b;    scanf("%d",&t);    while(t--){        getchar();        scanf("%c",&ch);        if(ch=='M'){            scanf("%d%d",&a,&b);            Move(a,b);        }        else if(ch=='C'){            scanf("%d",&a);            Find(a);//找到a的下面的木块            printf("%d\n",numunder[a]);        }    }    return 0;}