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hdu3635dragonballs（带权并查集）

hyp1231  hyp1231  2022-09-06  386

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=3635

题目描述：

Dragon Balls

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.
技术分享

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

题目大意：

　　i 球开始在 i 城市，然后现在 T A B 表示把A球和同城市的球移动到B球所在的城市

　　Q A 表示求A所在的城市、A所在的城市有几个球、A被移动了几次

思路：

　　想并查集的森林表示，移动了多少次实际就是到树根的距离（默认两点间的距离为1）

　　用三个数组记录状态

　　pre[i] 表示i球所在城市

　　num[i] 表示i城市有几个球

　　bt[i] 表示i球被移动了几次（就是i到树根的距离）

　　然后每次合并时让pre[ra] = rb 并 bt[ra] = 1

代码：

 1 #include <cstdio>
 2 using namespace std;
 3 
 4 const int N = 10010;
 5 
 6 int n, q, pre[N], num[N], bt[N];
 7 //pre[i] 表示i球所在城市
 8 //num[i] 表示i城市有几个球
 9 //bt[i] 表示i球被移动了几次（就是i到树根的距离）
10 
11 void init(int cas) {
12     printf("Case %d:
", cas);
13     for (int i = 1; i <= n; ++i)
14         pre[i] = i, num[i] = 1, bt[i] = 0;
15 }
16 
17 int find(int a) {
18     if (pre[a] == a)return a;
19     int tmp = pre[a];
20     pre[a] = find(pre[a]);
21     bt[a] = bt[a] + bt[tmp];
22     return pre[a];
23 }
24 
25 void merge(int a, int b) {
26     int ra = find(a), rb = find(b);
27     pre[ra] = rb, bt[ra] = 1;    //ra被移动了一次
28     num[rb] += num[ra], num[ra] = 0;
29     find(a);
30 }
31 
32 int main() {
33     int t, a, b;
34     scanf("%d", &t);
35     for (int o = 1; o <= t; ++o) {
36         scanf("%d%d", &n, &q);
37         init(o);
38         char tmp[2];
39         for (int i = 0; i < q; ++i) {
40             scanf("%1s", tmp);
41             if (tmp[0] == ‘T‘) {
42                 scanf("%d%d", &a, &b);
43                 merge(a, b);
44             } else {
45                 scanf("%d", &a);
46                 int ra = find(a);
47                 printf("%d %d %d
", ra, num[ra], bt[a]);
48             }
49         }
50     }
51 }

Dragon Balls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6373 Accepted Submission(s): 2398

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.
技术分享

Input

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

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