POJ 1988 -- Cube Stacking

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

http://blog.tpircsboy.com/acm/poj-1988/

除了parent数组，还要开设 under数组，under[i]表示第i个方块下面有多少个方块。 under数组在堆合并和路径压缩的时候都要更新。

const int maxn = 30010;

int parent[maxn],under[maxn];

 

int find(int t)

{

    if(parent[t]<0)

        return t;

    int s=find(parent[t]);

    under[t]+=under[parent[t]];

    parent[t]=s;

    return s;

}

void merge(int l,int r)

{

    int u = find(l);

    int d = find(r);

    if(u==d)

        return;

    under[u]=-parent[d];

    parent[d]+=parent[u];//将子集成员个数加起来

    parent[u]=d;

}

int main()

{

    freopen("in.txt","r",stdin);

    int p,l,r,t;

    char c,str[10];

    memset(parent,-1,sizeof(parent));

    scanf("%d",&p);

    while(p--)

    {

        //getchar();

        scanf("%s",str);

        if(str[0]=='M')

        {

            scanf("%d%d",&l,&r);

            merge(l,r);

        }

        if(str[0]=='C')

        {

            scanf("%d",&t);

            find(t);//输出前必须更新一下under数组

            printf("%d\n", under[t]);

        }

    }

}

http://www.cnblogs.com/Rinyo/archive/2013/02/23/2923686.html

一开始若干个元素自己为一个栈，给出n个操作，有如下两种：

　　　　1.M a b ：表示把元素a所在的栈整个压在含有元素b的栈的顶端

　　　　2.C x ：查询元素x所在的栈，x下方有几个元素，输出

　　题意简单明了：并查集

　　除了数组f[i]用来记录i的祖先，也就是顶端元素

　　另需要数组rank[i]，记录i所在的栈一共有多少个元素(i为栈顶)

　　　　   数组up[i]，记录i上面有多少个元素

　　则答案为rank[find(x)]-up[x]-1

至于在find和union中如何维护rank和up：

　　　　在find中只需维护up,即up[x]+=up[father[x]];

　　　　在union中需要维护up和rank，这时候rank起作用了。

　　　　例如将x所在的栈压在y所在栈顶端，则y所在栈的顶端father[y]上元素的个数即为x所在栈所有元素个数，即up[father[y]]=rank[father[x]]

　　　　（看了好多题解 表示不明白他们为什么要写up[father[y]]+=rank[father[x]]，个人觉得用不着+啊，而且+不+都能AC...躺倒）

　　　　而将x所在的栈压在y所在栈顶端后，x所在栈的总元素个数将会加上y所在栈元素个数，即rank[father[x]]+=rank[father[y]

http://www.cnblogs.com/lvpengms/archive/2010/02/03/1662792.html
* 应用 并查集的思路，移动一个栈时，相当于union_set操作，只要另开一个
* 数组记录立方体的位置，当合并时，只要改变根节点的位置记录就可以了，
* 这个地方并查集用的比较巧妙。其余的就是基本的并查集操作了
int cube[30330],nr[30330];int n;void make_set()
{
    for(int i=0;i<=30030;++i)
    {
        cube[i] = -1;
        nr[i] = 0;
    }
}int find_set(int x)
{
    int tmp = cube[x];
    if(cube[x]<0)
        return x;
    cube[x] = find_set(cube[x]);
    nr[x] += nr[tmp];
    return cube[x];
}void union_set(const int& root1,const int& root2)
{
    int tmp = cube[root1];
    cube[root1] = root2;
    nr[root1] =nr[root1] - cube[root2];
    cube[root2] += tmp;
}int main()
{
    int i,j,root1,root2,n1,n2;
    char ch[1];
    cin>>n;
    make_set();
    while(n--)
    {
        scanf("%s ",ch);  
        if(ch[0]=='M')
        {
            scanf("%d%d",&n1,&n2);
            int root1 = find_set(n1);
            int root2 = find_set(n2);
            if(root1!=root2)
                union_set(root1,root2);
        }else if(ch[0]=='C')
        {
            scanf("%d",&n1);
            int tmp =find_set(n1);
            printf("%d\n",nr[n1]);
        }
    }
    return 0;
}

Also check http://www.csdn123.com/html/blogs/20130522/15411.htm
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