POJ 1988 Cube Stacking(并查集)两种解法

　　　　地址：http://poj.org/problem?id=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

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

Sample Output

1
0
2
　　　　
　　　　题意：n个操作。M x y 表示将x所在的那堆箱子放在y那堆箱子头上。C x 询问x下面有多少箱子。
　　　解析一：以deep[x]记录x上面有多少箱子，堆顶为根，以all[x]记录x所在堆有多少个箱子。初始化为deep[]=0,all[]=1,pr[i]=i。find()函数的更新过程（包含路径压缩）

int find(int x)
{
    if(pr[x]==x)return x;
    int mid=pr[x];
    pr[x]=find(pr[x]);//路径压缩
    deep[x]+=deep[mid]; //下图记录此步
    return pr[x];
}

　　　　分析过程：

 　　　　接下来是join():

void join(int a,int b)
{
    int fa=find(a),fb=find(b);
    if(fa!=fb)
    {
        pr[fb]=fa;//以堆顶为根，a放在了b的头上
        deep[fb]+=all[fa];//下图分析此过程
        all[fa]+=all[fb];//下图分析此过程
    }
    return ;
}

　　　　 分析过程：

 　　　　计算当前点下面有多少个箱子，就是：all[find(x)]-deep[x]-1

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=3e5+50;
int n;
int pr[maxn],deep[maxn],all[maxn];
void init()
{
    for(int i=1;i<=n;i++)
    {
        pr[i]=i;
        deep[i]=0;
        all[i]=1;
    }
}
int find(int x)
{
    if(pr[x]==x)return x;
    int mid=pr[x];
    pr[x]=find(pr[x]);
    deep[x]+=deep[mid];
    return pr[x];
}
void join(int a,int b)
{
    int fa=find(a),fb=find(b);
    if(fa!=fb)
    {
        pr[fb]=fa;
        deep[fb]+=all[fa];
        all[fa]+=all[fb];
    }
    return ;
}
int main()
{
    while(~scanf("%d",&n))
    {
        init();
        char ch;
        int mid=n;
        while(mid--)
        {
            scanf(" %c",&ch);
            if(ch=='M')
            {
                int a,b;
                scanf("%d%d",&a,&b);
                join(a,b);
            }
            else
            {
                int x;
                scanf("%d",&x);
                cout<<all[find(x)]-deep[x]-1<<endl;
            }
        }
    }
}

　　　　解析二：以deep[x]记录x下面有多少箱子，堆底为根。其他与上面基本相同。find()相同。不同的是join()谁接谁改了一下：

void join(int a,int b)
{
    int fa=find(a),fb=find(b);
    if(fa!=fb)
    {
        pr[fa]=fb;//堆底为根。
        deep[fa]+=all[fb];//与上面不同
        all[fb]+=all[fa];//与上面不同
    }
    return ;
}

　　　　询问x，那么find(x)一下，防止之前没更新，直接输出deep[x]即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=3e5+50;
int n;
int pr[maxn],deep[maxn],all[maxn];
void init()
{
    for(int i=1;i<=n;i++)
    {
        pr[i]=i;
        deep[i]=0;
        all[i]=1;
    }
}
int find(int x)
{
    if(pr[x]==x)return x;
    int mid=pr[x];
    pr[x]=find(pr[x]);
    deep[x]+=deep[mid];
    return pr[x];
}
void join(int a,int b)
{
    int fa=find(a),fb=find(b);
    if(fa!=fb)
    {
        pr[fa]=fb;
        deep[fa]+=all[fb];
        all[fb]+=all[fa];
    }
    return ;
}
int main()
{
    while(~scanf("%d",&n))
    {
        init();
        char ch;
        int mid=n;
        while(mid--)
        {
            scanf(" %c",&ch);
            if(ch=='M')
            {
                int a,b;
                scanf("%d%d",&a,&b);
                join(a,b);
            }
            else
            {
                int x;
                scanf("%d",&x);
                find(x);
                cout<<deep[x]<<endl;
            }
        }
    }
}