poj1988（并查集）

最新推荐文章于 2019-04-29 22:56:12 发布

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本文介绍了一个名为CubeStacking的游戏，该游戏涉及堆叠立方体并执行移动和计数操作。玩家需要通过移动立方体堆来完成挑战，并计算特定立方体下方的立方体数量。文章提供了一段代码，用于验证游戏的操作结果。

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Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 26476		Accepted: 9264
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

#include<iostream>
#include<cstdio>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<list>
#include<set>
#include<iomanip>
#include<cstring>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cassert>
#include<sstream>
#include<algorithm>
using namespace std;
const int MAXN = 3e4+5;
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
typedef long long ll;

int n,fa[MAXN],r[MAXN],mx[MAXN];
//r[x]表示x到根节点的距离，mx[x]表示这个集合中有多少个盒子(集合大小)

void init()
{
	for(int i=1; i<MAXN; i++) fa[i]=i,r[i]=0,mx[i]=1;
}

int find(int x)
{
	int fx=fa[x];
	if(fa[x]!=x)
	{
		fx=find(fa[x]);
		r[x]+=r[fa[x]];
	}
	return fa[x]=fx;//返回祖先（路径压缩）
}

void U(int x,int y)//x集合放在y集合上面
{
	int fx=find(x),fy=find(y);
	fa[fy]=fx;
	r[fy]+=mx[fx];//y到根节点距离加长
	mx[fx]+=mx[fy];//x集合变大
}

int main()
{
	char op[3];
	int i,j;
	while(~scanf("%d",&n))
	{
		init();
		while(n--)
		{
			scanf("%s",op);
			if(op[0]=='C')
			{
				scanf("%d",&i);
				int f=find(i);
				printf("%d\n",mx[f]-r[i]-1);//集合盒子个数减去i盒子之上的个数
			}
			else
			{
				scanf("%d%d",&i,&j);
				U(i,j);
			}
		}
	}
	return 0;
}