Difference between revisions of "XX Open Cup named after E.V. Pankratiev. Grand Prix of Southeastern Europe"

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(Created page with "== Problem A == Unsolved. == Problem B == Upsolved by Kilo_5723. (-9) == Problem C == Unsolved. == Problem D == Solved by Kilo_5723. 01:38 (+6) == Problem E == Unsolv...")
 
Line 38: Line 38:
  
 
Solved by Xiejiadong. 01:56 (+)
 
Solved by Xiejiadong. 01:56 (+)
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题意:给一个完全图,可以讲图分割成好几个回路。回路经过的每一个点都有代价,代价是连接这个点的两条边中的较大值,现在可以安排回路使得回路代价的总和最小。
 +
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题解:可以发现对于每一个点,一定会有偶数次进出,一半的次数在进,一半的次数在出。
 +
 +
我们只需要决定了每一个节点的进出,因为是完全图,一定可以找到一种合法的回路。
 +
 +
于是问题就可以针对每一个点单独考虑。一进一出取最大值,最好的方案肯定是排序以后,从小到大两两组合。
 +
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于是针对每一个节点的权值排序计算即可。
  
 
== Problem K ==
 
== Problem K ==
  
 
Unsolved.
 
Unsolved.

Revision as of 06:26, 3 December 2019

Problem A

Unsolved.

Problem B

Upsolved by Kilo_5723. (-9)

Problem C

Unsolved.

Problem D

Solved by Kilo_5723. 01:38 (+6)

Problem E

Unsolved.

Problem F

Unsolved. (-3)

Problem G

Solved by Xiejiadong. 02:41 (+1)

Problem H

Unsolved.

Problem I

Solved by Kilo_5723. 00:32 (+2)

Problem J

Solved by Xiejiadong. 01:56 (+)

题意:给一个完全图,可以讲图分割成好几个回路。回路经过的每一个点都有代价,代价是连接这个点的两条边中的较大值,现在可以安排回路使得回路代价的总和最小。

题解:可以发现对于每一个点,一定会有偶数次进出,一半的次数在进,一半的次数在出。

我们只需要决定了每一个节点的进出,因为是完全图,一定可以找到一种合法的回路。

于是问题就可以针对每一个点单独考虑。一进一出取最大值,最好的方案肯定是排序以后,从小到大两两组合。

于是针对每一个节点的权值排序计算即可。

Problem K

Unsolved.