×_7_12_2013 C: More happiness

最新推荐文章于 2025-09-10 17:59:49 发布

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最新推荐文章于 2025-09-10 17:59:49 发布

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文章标签： python

原文链接：https://my.oschina.net/dianpaopao/blog/145703

本文提供了一种方法，帮助计划在 Happy Valley 旅行时最大化游览景点的数量，考虑到每处景点的行走时间和总可停留时间。通过输入旅行的景点数量、总时间限制和每处景点的猜测停留时间，输出可以游览的最大景点数量。

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1 Solved: 0
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As you know, poor Xiaoming has been dreaming to go to Happyvalley for a long time. The good news is

that today his mom finally agrees to take him there this weekend!

In order to make his journey in Happyvalley happier, Xiaoming found a map of Happyvalley to design a

good path. Happyvalley has N scenic spots and there are roads between them. Xiaoming found that there is

only one way between each two spots. In other words, the N spots and roads between them form a “tree”.

Xiaoming can start his journey from arbitrary scenic spot. It takes Ci minutes to walk from one end to the

other end on the ith road. Xiaoming knows that he can only stay in Happyvalley for only P minutes.

However, he is uncertain of the time he may spend in each scenic spot. So he can only guess it.

Given the information of Happyvalley and the total time P minutes, also, with the time X Xiaoming guess he

would spend in each spot, can you tell Xiaoming the maximum number of spots he can stay for X minutes?

Note that each spot should be count only once.

The first line of the input is T (no more than 10), which stands for the number of test cases you need to

solve.

On the first line of each test case, there are two integers giving the number N (no more than 200) of scenic

spots in Happyvalley and the total time P (no more than 2000000) Xiaoming can stay. The following N-1

lines state the roads in the form of “a b c” which means it takes c minutes to walk from spot a to spot b or

from b to a (1 <= a, b <= N, 1 <= c <= 10000).

Then an integer Q (no more than 10000) in a single line is the number of queries Xiaoming asks you. Each

of the next N lines gives a number X (no more than 10000) representing the time Xiaoming guess.

Output the maximum number of spots he can get based on each guess of Xiaoming. Separate each test case

with a blank line.

2 3 10 1 2 2 2 3 1 3 10 2 4 6 15 1 2 2 1 3 1 2 6 3 5 3 5 3 4 3 4 1 2 3 100

1 3 2 5 4 3 0

转载于:https://my.oschina.net/dianpaopao/blog/145703