poj Cube Stacking（带权并查集）

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最新推荐文章于 2022-07-05 13:10:43 发布

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分类专栏： # 并查集 ACM 文章标签： poj Cube Stacking

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本文介绍了一个名为CubeStacking的游戏模拟问题，该游戏涉及N个相同的方块，玩家需要执行一系列移动和计数操作。通过使用带权并查集的数据结构，文章详细解释了如何高效地实现这些操作，并提供了一段完整的C语言程序代码来验证游戏的结果。

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Cube Stacking

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

Sample Output

1
0
2

题意：

n个方块，p组操作。操作有两种，M a b 是将含有a的堆放在含有b的堆上，C a 是输出在含有a的堆里a下面共有多少个方块

思路：

带权并查集，让一个堆中最上面的方块为父节点，dis[a]代表a到父节点的距离，rank[pre[a]]代表含有a的堆的大小，那么答案即为rank[pre[a]]-dis[a]-1

代码：

#include<stdio.h>
const int maxn=30000+10;
int pre[maxn],dis[maxn],rank[maxn];
void init()
{
    for(int i=0;i<maxn;++i)
    {
        pre[i]=i;
        dis[i]=0;
        rank[i]=1;
    }
}
int Find(int x)
{
    if(x!=pre[x])
    {
        int t=pre[x];
        pre[x]=Find(pre[x]);
        dis[x]+=dis[t];
    }
    return pre[x];
}
void join(int x,int y)
{
    int fx=Find(x),fy=Find(y);
    if(fx!=fy)
    {
        pre[fy]=fx;
        dis[fy]=rank[fx];
        rank[fx]+=rank[fy];
    }
}
int main()
{
    int i,j,n;
    while(~scanf("%d",&n))
    {
        init();
        char op;
        for(i=1;i<=n;++i)
        {
            getchar();
            int a,b;
            scanf("%c",&op);
            if(op=='M')
            {
                scanf("%d%d",&a,&b);
                join(a,b);
            }
            else
            {
                scanf("%d",&a);
                int x=Find(a);
                printf("%d\n",rank[x]-dis[a]-1);
            }
        }
    }
    return 0;
}