#Lutece1219. Ba Gua Zhen
Ba Gua Zhen
Migrated from Lutece 1219 Ba Gua Zhen
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Description
During the Three-Kingdom period, there was a general named Xun Lu who belonged to Kingdom Wu. Once his troop were chasing Bei Liu, he was stuck in the Ba Gua Zhen from Liang Zhuge. The puzzle could be considered as an undirected graph with vertexes and edges. Each edge in the puzzle connected two vertexes which were and with a length of . Liang Zhuge had great interests in the beauty of his puzzle, so there were no self-loops and between each pair of vertexes, there would be at most one edge in the puzzle. And it was also guaranteed that there was at least one path to go between each pair of vertexes.
Fortunately, there was an old man named Chengyan Huang who was willing to help Xun Lu to hack the puzzle. Chengyan told Xun Lu that he had to choose a vertex as the start point,
then walk through some of the edges and return to the start point at last. During his walk, he could go through some edges any times. Since Liang Zhuge had some mysterious magic,
Xun Lu could hack the puzzle if and only if he could find such a path with the maximum XOR
sum of all the edges length he has passed. If the he passed some edge multiple times,
the length would also be calculated by multiple times. Now, could you tell Xun Lu which is the maximum XOR
circuit path in this puzzle to help him hack the puzzle?
Input
The first line of the input gives the number of test cases, (). test cases follow.
Each test case begins with two integers () and () in one line. Then lines follow. Each line contains three integers , and (, ) to describe all the edges in the puzzle.
Output
For each test case, output one line containing Case #x: y
, where is the test case number (starting from ) and is the maximum XOR
sum of one circuit
path in the puzzle.
Samples
2
3 3
1 2 1
1 3 2
2 3 0
6 7
1 2 1
1 3 1
2 3 1
3 4 4
4 5 2
4 6 2
5 6 2
Case #1: 3
Case #2: 3
Note
A XOR
takes two bit patterns of equal length and performs the logical exclusive OR
operation on each pair of corresponding bits. The result in each position is
if only the first bit is or only the second bit is , but will be if both are or both are . In this we perform the comparison of two bits, being if the
two bits are different, and if they are the same.
Resources
The 2015 China Collegiate Programming Contest