hdoj 3635 Dragon Balls【并查集】

最新推荐文章于 2018-12-23 14:25:23 发布

坚定的水手

最新推荐文章于 2018-12-23 14:25:23 发布

阅读量391

点赞数 1

CC 4.0 BY-SA版权

文章标签：并查集

本文链接：https://blog.youkuaiyun.com/liu6886/article/details/47207233

本文介绍了一个基于并查集的数据结构算法问题，通过该算法可以高效地解决龙珠位置追踪及转移次数统计的问题。该问题设定在一个包含多个城市的环境中，每个城市可能存放有一个或多个龙珠，随着时间推移，龙珠会在不同城市间转移。算法需要实现的功能包括记录每次转移后的最新位置，并能够快速查询任意指定龙珠的位置、所在城市龙珠总数及其转移次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4364 Accepted Submission(s): 1667

Problem Description

Five hundred years later, the number of dragon balls will increase unexpectedly, so it's too difficult for Monkey King(WuKong) to gather all of the dragon balls together.

His country has N cities and there are exactly N dragon balls in the world. At first, for the ith dragon ball, the sacred dragon will puts it in the ith city. Through long years, some cities' dragon ball(s) would be transported to other cities. To save physical strength WuKong plans to take Flying Nimbus Cloud, a magical flying cloud to gather dragon balls.
Every time WuKong will collect the information of one dragon ball, he will ask you the information of that ball. You must tell him which city the ball is located and how many dragon balls are there in that city, you also need to tell him how many times the ball has been transported so far.

Input

The first line of the input is a single positive integer T(0 < T <= 100).
For each case, the first line contains two integers: N and Q (2 < N <= 10000 , 2 < Q <= 10000).
Each of the following Q lines contains either a fact or a question as the follow format:
T A B : All the dragon balls which are in the same city with A have been transported to the city the Bth ball in. You can assume that the two cities are different.
Q A : WuKong want to know X (the id of the city Ath ball is in), Y (the count of balls in Xth city) and Z (the tranporting times of the Ath ball). (1 <= A, B <= N)

Output

For each test case, output the test case number formated as sample output. Then for each query, output a line with three integers X Y Z saparated by a blank space.

Sample Input

Sample Output

Case 1:
2 3 0
Case 2:
2 2 1
3 3 2

Q A
X：A号球目前所在的 X城的ID （找父节点）
Y：X城目前的球总数（根节点总数）
Z：A号球被转移了几次（计数器数组）

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define  N 10010
using namespace std;
int n,m;
int f[N],r[N],t[N];//  f[N] 父节点   r[N] 根节点 t[N] 计数器 累计不同球转移次数 
void init()
{
	memset( t , 0 , sizeof(t));
	for( int i=0;i<=n;i++)
	{
		f[i]=i;
		r[i]=1;
	}
}
//查找 父节点  根节点  压缩路径  计算转移次数 
int find( int x )
{
	if( f[x] !=x)
	{
		int fa = f[x];
		f[x] = find(fa);
		r[x]+=r[fa];
		t[x]+=t[fa];
	}
	return f[x];
}
// 合并  父节点  
void join( int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if( fx!= fy)
	{
		f[fx] = fy;
		r[fy] +=r[fx];
		t[fx]++;
	}
}
void solve(int x)
{
	int fx=find(x);
	printf("%d %d %d\n",fx,r[fx],t[x]);
 } 
 int main()
 {
 	int h,ca=1;
 	char c;
 	int x,y;
 	scanf("%d",&h);
 	while(h--)
 	{	 	
 		scanf("%d%d",&n,&m);
 		getchar();
 		init();		 
		printf("Case %d:\n",ca++);	
		while(m--)
		{
			c = getchar();
			if(c=='T')
			{
				scanf("%d%d",&x,&y);
				getchar();
				join(x,y);
			}
			else
			{
				scanf("%d",&x);
				getchar();
				solve(x);
			}
		} 
	}
 	return 0;
}