HDU 3635 Dragon Balls

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1419    Accepted Submission(s): 565

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

  

   2
3 3
T 1 2
T 3 2
Q 2
3 4
T 1 2
Q 1
T 1 3
Q 1
  

 

Sample Output

  

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

 

Author

possessor WC

 

Source

2010 ACM-ICPC Multi-University Training Contest（19）——Host by HDU

 

Recommend

lcy
解题思路：依然是并查集的题目，难点是记录每个龙珠的移动次数，首先若移动A及A所在城市的所有龙珠移动到B龙珠所在城市时，A所在路径上的最远祖先一定会移动一次，然后在Q查询时，从最远祖先之前的那个龙珠开始，把移动步数加上其压缩成最短路径之前的祖先（由temp记录）的移动步数就是移动到其最远祖先所需要的步数。

#include <iostream>
#include<cstring>
using namespace std;
int father[10010],num[10010],Move[10010];
void initial()
{
    int i;
    for(i=0;i<10010;i++)
    {
        father[i]=i;
        num[i]=1;
    }
	memset(Move,0,sizeof(Move));
}
int find(int a)
{
	int temp;
    if(a==father[a])
		return a;
    temp=father[a];//temp会记录下路径压缩前每个节点的祖先节点
	father[a]=find(father[a]);
	Move[a]+=Move[temp];//把路径上的每个龙珠的移动步数加上temp的移动步数
	return father[a];
}
int main()
{
    int i,n,t=1,city,op,ball,ball1,ball2,a,b,c;
    char ch[2];
    scanf("%d",&n);
    while(n--)
    {
        initial();
        scanf("%d%d",&city,&op);
		printf("Case %d:\n",t);
				t++;
        for(i=0;i<op;i++)
        {
          scanf("%s",&ch);
        if(ch[0]=='T')
        {
            scanf("%d%d",&ball1,&ball2);
            int x=find(ball1);
            int y=find(ball2);       
            if(x!=y)
            {
			   Move[x]++;
               father[x]=y;
               num[y]+=num[x];
            }
        }
        else
        {
            scanf("%d",&ball);
            a=find(ball);
            b=num[a];
            c=Move[ball];
            printf("%d %d %d\n",a,b,c);
        }
        }
    }
    return 0;
}