2020-2021:teams:alchemist:2020_self_training_1
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2020-2021:teams:alchemist:2020_self_training_1 [2020/07/23 22:22] mountvoom [I - Snake Carpet] |
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 ===== | ||
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| + | 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 | ||
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| + | ===== D - Kejin Game ===== | ||
| + | |||
| + | 一个网络流题目，建图方式和最大权闭合图有点像 | ||
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| + | $source \rightarrow i$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 | ||
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| + | by Hardict | ||
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| + | ===== G - Mysterious Antiques in Sackler Museum ===== | ||
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| + | $4$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 | ||
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| + | by Hardict | ||
| ===== I - Snake Carpet ===== | ===== I - Snake Carpet ===== | ||
| 行 71: | 行 97: | ||
| by MountVoom | by MountVoom | ||
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| + | ===== K - A Math Problem ===== | ||
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| + | 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$dp$ | ||
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| + | by Hardict | ||
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2020-2021/teams/alchemist/2020_self_training_1.1595514126.txt.gz · 最后更改: 2020/07/23 22:22 由 mountvoom
