Difference between revisions of "2019 Multi-University,Nowcoder Day 5"
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Xiejiadong (talk | contribs) (Created page with "== Problem A == Solved by Kilo_5723. 00:17:42 (+) == Problem B == Solved by Kilo_5723. 01:27:34 (+) == Problem C == Unsolved. (-8) == Problem D == Unsolved. == Problem...") |
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Solved by Xiejiadong. 03:00:20 (+2) | Solved by Xiejiadong. 03:00:20 (+2) | ||
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| + | 题意：每次给出两个字母的相对关系，还原原字符串。 | ||
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| + | 题解：每次给出的字母一定是包含了所有出现位置的，否则就是无解的情况。 | ||
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| + | 于是我们先对所有的字母编号，然后在每次给出的顺序相邻之间连边（所有边都连的话，可能会 TLE ）。 | ||
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| + | 在连出的图中跑一个拓扑排序就好了。 | ||
== Problem I == | == Problem I == | ||
Revision as of 11:40, 1 August 2019
Problem A
Solved by Kilo_5723. 00:17:42 (+)
Problem B
Solved by Kilo_5723. 01:27:34 (+)
Problem C
Unsolved. (-8)
Problem D
Unsolved.
Problem E
Unsolved.
Problem F
Unsolved. (-15)
Problem G
Solved by Xiejiadong. 03:42:19 (+)
Problem H
Solved by Xiejiadong. 03:00:20 (+2)
题意：每次给出两个字母的相对关系，还原原字符串。
题解：每次给出的字母一定是包含了所有出现位置的，否则就是无解的情况。
于是我们先对所有的字母编号，然后在每次给出的顺序相邻之间连边（所有边都连的话，可能会 TLE ）。
在连出的图中跑一个拓扑排序就好了。
Problem I
Solved by Kilo_5723. 03:13:09 (+1)
Problem J
Unsolved. (-12)