Difference between revisions of "XX Open Cup named after E.V. Pankratiev. Grand Prix of Southeastern Europe"
Jump to navigation
Jump to search
Xiejiadong (talk | contribs) (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...") |
Xiejiadong (talk | contribs) |
||
Line 38: | Line 38: | ||
Solved by Xiejiadong. 01:56 (+) | 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 == | == 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.