Dragon Balls
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7926 Accepted Submission(s): 2937
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.
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)
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
题目意思：
给你n个物品，从1到n编号
现在对n有m个操作
T A B 把A所在集合的物品全部移到B所在的集合（移动的物品包括A）
Q X 问你X所在集合的编号，x所在集合的结点数量，和x移到的次数
给你n个物品，从1到n编号
现在对n有m个操作
T A B 把A所在集合的物品全部移到B所在的集合（移动的物品包括A）
Q X 问你X所在集合的编号，x所在集合的结点数量，和x移到的次数
注意：
比如样例1：1移到2，然后3移到2，因为是移到2，所以集合编号就是2，注意理解
此时Q 1，那么1所在集合编号就是2，因为是移到2去的，1所在集合的结点数量是3
那么此时1的移到此时是1
比如样例1：1移到2，然后3移到2，因为是移到2，所以集合编号就是2，注意理解
此时Q 1，那么1所在集合编号就是2，因为是移到2去的，1所在集合的结点数量是3
那么此时1的移到此时是1
注意理解移动次数数组
具体参考代码
#include<queue> #include<set> #include<cstdio> #include <iostream> #include<algorithm> #include<cstring> #include<cmath> #include<map> #include<string> #include<string.h> #include<memory> using namespace std; #define max_v 10005 #define INF 9999999 int pa[max_v]; int num[max_v];//num[i] i所在集合内含有结点个数 int mov[max_v];//mov[i] i移动的次数 int n,m; void init() { for(int i=0; i<=n; i++) { pa[i]=i; num[i]=1; mov[i]=0; } } int find_set(int x) { if(pa[x]!=x) { int t=pa[x]; pa[x]=find_set(pa[x]); mov[x]+=mov[t];//孩子结点的移动次数会与其父亲结点移动次数相关 } return pa[x]; } void union_set(int x,int y) { int fx=find_set(x); int fy=find_set(y); if(fx!=fy) { pa[fx]=fy; num[fy]+=num[fx];//合并之后大集合的结点数等于原来两个小集合结点数目之和 mov[fx]++;//x的根结点 移动次数++ x的移动次数在find_set函数里面更新 } } int main() { int t; int x,y; char str[10]; int c=1; scanf("%d",&t); while(t--) { scanf("%d %d",&n,&m); init(); printf("Case %d: ",c++); for(int i=0; i<m; i++) { scanf("%s",str); if(str[0]=='T') { scanf("%d %d",&x,&y); union_set(x,y); } else if(str[0]=='Q') { scanf("%d",&x); int k=find_set(x); printf("%d %d %d ",k,num[k],mov[x]); } } } return 0; } /* 题目意思： 给你n个物品，从1到n编号 现在对n有m个操作 T A B 把A所在集合的物品全部移到B所在的集合（移动的物品包括A） Q X 问你X所在集合的编号，x所在集合的结点数量，和x移到的次数 注意： 比如样例1：1移到2，然后3移到2，因为是移到2，所以集合编号就是2，注意理解 此时Q 1，那么1所在集合编号就是2，因为是移到2去的，1所在集合的结点数量是3 那么此时1的移到此时是1 注意理解移动次数数组 具体参考代码 */