HOLI - Holiday Accommodation

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Nowadays, people have many ways to save money on accommodation when they are on vacation. One of these ways is exchanging houses with other people.

Here is a group of N people who want to travel around the world. They live in different cities, so they can travel to some other people's city and use someone's house temporary. Now they want to make a plan that choose a destination for each person. There are two rules should be satisfied:

  1. All the people should go to one of the other people's city.
  2. Two of them never go to the same city, because they are not willing to share a house.

They want to maximize the sum of all people's travel distance. The travel distance of a person is the distance between the city he lives in and the city he travels to. These N cities have N-1 highways connecting them. The travelers always choose the shortest path when traveling.

Given the highways' information, it is your job to find the best plan, that maximum the total travel distance of all people.

Input

The first line of input contains one integer T (1 ≤ T ≤ 10), indicating the number of test cases.

Each test case contains several lines. The first line contains an integer N (2 ≤ N ≤ 105), representing the number of cities. Then the following N-1 lines each contains three integers X, Y, Z (1 ≤ X, YN, 1 ≤ Z ≤ 106), means that there is a highway between city X and city Y, and length of that highway.

You can assume all the cities are connected and the highways are bi-directional.

Output

For each test case in the input, print one line: "Case #X: Y", where X is the test case number (starting with 1) and Y represents the largest total travel distance of all people.

Example

Input:
2
4
1 2 3
2 3 2
4 3 2
6
1 2 3
2 3 4
2 4 1
4 5 8
5 6 5

Output:
Case #1: 18
Case #2: 62

hide comments
iamjacksparrow: 2019-04-04 00:39:49

take visited array size 1e5+1( not 1e5 :-( )

ankit0107verma: 2019-01-03 13:39:35

Beware avoid using maps for visited and counting nodes in dfs ,use unordered map or array else you will see the wrath of TLE

ankitpriyarup: 2018-12-10 17:32:45

After 3 WA finally AC :)

riyuzaki251097: 2018-08-22 11:54:54

great question learnt a lot , hint is to use pigeonhole theorem

sahil070197: 2018-02-23 10:55:41

Last edit: 2018-02-23 10:56:23
m0j0_j0j0: 2018-02-23 10:55:26

AC in one GO :P

aruneshg: 2018-02-01 19:47:30

what is the significance of the tree type of graph ??

vivace: 2017-06-06 21:34:51

Very nice quest . Beware strict time limits. One of those quests that distinguish between O(f(n)) and O(f(n)*logn)).
The logic is really good.

Last edit: 2017-06-06 21:35:28
vivace: 2017-06-06 21:34:51

Very nice quest . Beware strict time limits. One of those quests that distinguish between O(f(n)) and O(f(n)*logn))


Added by:Bin Jin
Date:2011-11-08
Time limit:0.738s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:ACM/ICPC Regional Contest, Chengdu 2011