Difference between revisions of "2019 Multi-University,Nowcoder Day 2"
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Xiejiadong (talk | contribs) (Created page with "== Problem A == Solved by Weaver_zhu && Kilo_5723. 01:36:01 (+2) == Problem B == Unsolved. == Problem C == Unsolved. == Problem D == Unsolved. == Problem E == Unsolve...") |
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Solved by Xiejiadong. 03:43:42 (+2) | Solved by Xiejiadong. 03:43:42 (+2) | ||
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| + | 题意：求全 $1$ 的次大矩阵面积。 | ||
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| + | 题解：考虑最大矩阵如何做，对于每一行考虑这一行作为底的最大矩形面积。 | ||
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| + | 这个就变成了经典的最大广告牌问题，用单调栈维护每个位置向左向右最多能拓展的长度，即可算出最的矩形。 | ||
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| + | 对于次大的矩形，考虑裁掉最大矩形的四个角，在分别算最大的矩形面积，取这四个中的较大值即可。 | ||
| + | |||
| + | 脑子不动，暗示 Weaver_zhu 写了三个小时的三分，于是这题爆炸了。 | ||
== Problem I == | == Problem I == | ||
Revision as of 11:38, 20 July 2019
Problem A
Solved by Weaver_zhu && Kilo_5723. 01:36:01 (+2)
Problem B
Unsolved.
Problem C
Unsolved.
Problem D
Unsolved.
Problem E
Unsolved.
Problem F
Upsolved by Weaver_zhu. (-1)
Problem G
Unsolved.
Problem H
Solved by Xiejiadong. 03:43:42 (+2)
题意：求全 $1$ 的次大矩阵面积。
题解：考虑最大矩阵如何做，对于每一行考虑这一行作为底的最大矩形面积。
这个就变成了经典的最大广告牌问题，用单调栈维护每个位置向左向右最多能拓展的长度，即可算出最的矩形。
对于次大的矩形，考虑裁掉最大矩形的四个角，在分别算最大的矩形面积，取这四个中的较大值即可。
脑子不动，暗示 Weaver_zhu 写了三个小时的三分，于是这题爆炸了。
Problem I
Unsolved.
Problem J
Solved by Kilo_5723. 04:37:15 (+6)