POJ 1988 Cube Stacking 帶權並查集

Cube Stacking
Time Limit: 2000MS      Memory Limit: 30000K
Total Submissions: 23568        Accepted: 8258
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
Source
USACO 2004 U S Open

來源： http://poj.org/problem?id=1988

#include <cstdio>
using namespace std;
#define MAXN 30000+500
struct  LNode
{
    int father;
    int Len_root;       //到父親的距離
    int num;            //如果是根節點的話，集合元素個數
    int ancestor;       //祖先
}Animals[MAXN];
void Init(int n)
{
    for (int i = 1;i <= n;i++)
    {
        LNode &Ani = Animals[i];
        Ani.ancestor = Ani.father = i;
        Ani.Len_root = 0;   //距離自身爲0
        Ani.num = 0;        //後續元素個數爲0
    }
}
int Len(int x,int ance)
{
    if(x==ance) return 0;
    else  return Animals[x].Len_root + Len(Animals[x].father,ance);
}
int Find(int x)
{
    LNode &Ani = Animals[x];
    if (Ani.ancestor == x) return x;
    else
    {
        int Real_ance = Find(Ani.ancestor);
        Ani.ancestor  = Real_ance;                    //路徑壓縮1
        Ani.Len_root  = Len(x,Real_ance);
        Ani.father    = Real_ance;                     //路徑壓縮2
        Ani.num = Animals[Real_ance].num - Ani.Len_root;
        return Ani.ancestor;
    }
}
void Unite(int x,int y,int Rx_y = 1)  //Y->X
{
    int X = Find(x), Y = Find(y);
    if (X == Y) return;
    LNode &YL = Animals[Y];
    YL.father = YL.ancestor = X;
    LNode &XL = Animals[X];
    YL.Len_root = ++XL.num;
    XL.num += YL.num;
}
int main(void)
{
    //freopen("F:\\test.txt","r",stdin);
    int P;char None[200];
    scanf("%d",&P);
    gets(None);
    Init(30000);char order;
    for(int i=1,a,b;i<=P;i++)
    {
        scanf("%c",&order);
        if(order == 'M')
        {
            scanf("%d %d",&a,&b);
            Unite(a,b);
        }
        else
        {
            scanf("%d",&a);Find(a);
            printf("%d\n",Animals[a].num);
        }
        gets(None);
    }
}