HDU 3635 Dragon Balls

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 766 Accepted Submission(s): 308 

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

23 3T 1 2T 3 2Q 23 4T 1 2Q 1T 1 3Q 1

 

Sample Output

Case 1:2 3 0Case 2:2 2 13 3 2

 

 题目大意：

初始时，有n个龙珠，编号从1到n，分别对应的放在编号从1到n的城市中。

现在又2种操作：

T A B，表示把A球所在城市全部的龙珠全部转移到B城市。（第一次时，因为A球所在的城市只有一个球，所以只移动1个，如果有多个，则全部移动）。

Q A，表示查询A。要求得到的信息分别是：A现在所在的城市，A所在城市的龙珠数目，A转移到该城市移动的次数（如果没有移动就输出0）

思路：

对于 所在城市 就是所在根节点，所在城市的龙珠数也很好弄，加一个 num  处理就行了，最难弄的是次数，其实在路径压缩法里面，find 就相当于移动的次数。所以顺理成章，移动次数的问题就解决了。

mdzz  对于输出 case 的题目，每次都忘记 ++  

AC代码：

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;const int MAX = 10000 ;int parent[MAX+10];int mmove[MAX+10];int num[MAX+10];int GetParent(int a){    if( parent[a]== a)        return a;    int t =parent[a];    parent[a] = GetParent(parent[a]);    mmove[a]+=mmove[t];    return parent[a];}void Merge(int b,int a){    int p1 = GetParent(a);    int p2 = GetParent(b);    if( p1 == p2 )        return;    parent[p2] = p1;    mmove[p2]=1;    num[p1]+=num[p2];}int main(){    int T;    int n,m;    scanf("%d",&T);    int ccc=1;    while(T--)    {        cout<<"Case "<<ccc<<":"<<endl;        ccc++;        scanf("%d%d",&n,&m);        for(int i=0;i<=n;i++)        {            parent[i]=i;            num[i]=1;            mmove[i]=0;        }        int h,s;        char cc[10];        for(int i=0;i<m;i++)        {            scanf("%s",cc);            if(cc[0]=='T')            {                scanf("%d%d",&h,&s);                Merge(h,s);            }            else            {                scanf("%d",&h);                s=GetParent(h);                cout<<s<<" "<<num[s]<<" "<<mmove[h]<<endl;            }        }    }}