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2020-2021:teams:alchemist:2020_self_training_1 [2020/07/23 22:11] mountvoom [C - Today Is a Rainy Day] |
2020-2021:teams:alchemist:2020_self_training_1 [2020/07/24 10:38] (当前版本) hardict [lpy] |
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===== cmx ===== | ===== cmx ===== | ||
===== lpy ===== | ===== lpy ===== | ||
+ | |||
+ | 网络流多组数据初始化时得多加注意 | ||
===== xsy ===== | ===== xsy ===== | ||
行 32: | 行 34: | ||
by MountVoom | by MountVoom | ||
+ | |||
+ | ===== D - Kejin Game ===== | ||
+ | |||
+ | 一个网络流题目,建图方式和最大权闭合图有点像 | ||
+ | |||
+ | $source \rightarrow i$流量为正常学习花费 | ||
+ | |||
+ | $i \rightarrow i'$流量为氪金学习花费 | ||
+ | |||
+ | $i' \rightarrow j$流量为删除$j$对$i$依赖价值 | ||
+ | |||
+ | 而最关键的是$target \rightarrow sink$为一条$\infty$的边 | ||
+ | |||
+ | 则原图每个**有限**割都对应一种学习发案,求最小花费就是求最小割 | ||
+ | |||
+ | by Hardict | ||
+ | |||
+ | ===== G - Mysterious Antiques in Sackler Museum ===== | ||
+ | |||
+ | $4$个矩形,判断能否选$3$个出来拼接为一个更大的矩形 | ||
+ | |||
+ | 暴力枚举即可,可以利用$next_permutation$减少码量 | ||
+ | |||
+ | by Hardict | ||
===== I - Snake Carpet ===== | ===== I - Snake Carpet ===== | ||
构造题。当时手动构造出来了(震惊),当时发现奇数能很好的堆成一个块,然后发现偶数又能恰好在旁边堆上2333。 | 构造题。当时手动构造出来了(震惊),当时发现奇数能很好的堆成一个块,然后发现偶数又能恰好在旁边堆上2333。 | ||
+ | <code c> | ||
+ | 1 | ||
+ | 22 | ||
+ | 332 | ||
+ | 132 | ||
+ | 33244 | ||
+ | 13244 | ||
+ | 55544 | ||
+ | 33544 | ||
+ | 13522 | ||
+ | |||
+ | 5554466 | ||
+ | 3354466 | ||
+ | 1352266 | ||
+ | |||
+ | 7777666 | ||
+ | 5557666 | ||
+ | 3357244 | ||
+ | 1357244 | ||
+ | |||
+ | 777766688 | ||
+ | 555766688 | ||
+ | 335724488 | ||
+ | 135724488 | ||
+ | </code> | ||
+ | |||
+ | 看了这个表就很清晰了,如果$n$是偶数考虑$n - 1$最后再加两列$n$即可。 | ||
+ | |||
+ | 构造奇数时发现$n$可以很容易的由$n - 4$转移过来,于是问题就解决啦。 | ||
by MountVoom | by MountVoom | ||
+ | |||
+ | ===== K - A Math Problem ===== | ||
+ | |||
+ | 打表发现$f(2x)=3f(x),f(2x+1)=3f(x)+1$ | ||
+ | |||
+ | 然后可以发现$f(x)$的$3$进制表示数值上等于$x$的$2$进制表示 | ||
+ | |||
+ | 其实就是个取模带余数的数位$dp$ | ||
+ | |||
+ | by Hardict | ||
+ |