POJ - 1988-（带权并查集）

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

题意：M    a    b，将a所在的方块整体移动到b所在的方块上。C    a查询a方块下面的方块数。

题解：不能简单的套用并查模板，维护一个数组记录到根距离，在合并·的时候需要不断更新。

还需另开一数组统计a,b所在的方块数。

详细代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int m;
int f[30500];
int deep[30500];          //记录方块到根的距离
int num[30500];           //记录此时所在的方块数量
void init()
{
    for(int i=0; i<=30001; i++)
    {
        f[i]=i;
        deep[i]=0;        //
        num[i]=1;         //初始时为一啦
    }
}
//找祖宗，并更新距离
int getf(int v)
{
    if(f[v]==v)return v;
    int fa=f[v];         // 
    f[v]=getf(f[v]);     //递归找祖宗
    deep[v]+=deep[fa];   //通过递归不断地更新距离（V与父亲距离+父亲与父亲的父亲+....与·祖宗的距离）
    return f[v];
}
//合并
void merge(int v,int u)
{
    int t1=getf(v);
    int t2=getf(u);
    if(t1!=t2)
    {
        f[t1]=t2;
        deep[t1]+=num[t2];  //合并时要加上方块数
        num[t2]+=num[t1];   
        num[t1]+=num[t2];
    }
}
int main()
{
    while(~scanf("%d",&m))
    {
        char s[5];
        init();
        int a,b;
        for(int i=0; i<m; i++)
        {
            scanf("%s",&s);
            if(strcmp(s,"M")==0)
            {
                scanf("%d%d",&a,&b);
                merge(a,b);
            }
            else
            {
                scanf("%d",&a);
                getf(a);  //这一步很重要，因为在此之前可能没有更新他的距离（先前没有输入过）
                printf("%d\n",deep[a]);
            }
        }
    }
    return 0;
}