Cube Stacking（并查集）-优快云博客

本文介绍了一款名为CubeStacking的游戏，该游戏涉及堆叠立方体的操作，并通过移动和计数两种操作来完成挑战。文章提供了详细的算法实现，包括并查集的应用及代码示例。

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http://poj.org/problem?id=1988

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 22312		Accepted: 7823
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

Sample Output

1
0
2

Source

USACO 2004 U S Open

简单并查集，入门好题

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int SIZE=3e4+10;
int parent[SIZE];
int under[SIZE],sum[SIZE];
int getroot(int x){
    if(parent[x]==x)return x;
    int fx=getroot(parent[x]);//递归获得根，前面的层数的返回最后的根节点
    under[x]+=under[parent[x]];//是parent[x]不是fx，给的例子中先最新的fx的under是0，而加的应该是原先的under值,
    return parent[x]=fx;
}//一次性把一条路径上的点更新了，每个parent[x]都是保存了当初一次的under的值，不太好表达
void Merge(int x,int y){
    int fx=getroot(x),fy=getroot(y);
    if(fx==fy)return ;
    parent[fx]=fy;
    under[fx]=sum[fy];
    sum[fy]+=sum[fx];
}
int main()
{
    int p,x,y,c;
    while(scanf("%d",&p)!=EOF){
        char s[10];
        for(int i=1;i<SIZE;i++){
            under[i]=0;
            sum[i]=1;
            parent[i]=i;
        }
        while(p--){
            scanf("%s",s);
            if(s[0]=='M'){
                scanf("%d%d",&x,&y);
                Merge(x,y);
            }
            else {
                scanf("%d",&c);
                getroot(c);
                printf("%d\n",under[c]);
            }
        }
        return 0;
    }
}
/**
M 1 2
M 1 3
M 1 4
画画图
***/