poi 17th(九) - HZH's Blog - 自闭症
poi 17th(九)
之所以挑这届当然也是因为比较水啦~
| Guilds | (Stage I) | (100/100) |
| Railway | (Stage I) | (0/100) |
| Beads | (Stage I) | (100/100) |
| Divine Divisor | (Stage I) | (0/100) |
| Intelligence Test | (Stage I) | (100/100) |
| Antisymmetry | (Stage II - day 0) | (100/100) |
| Hamsters | (Stage II - day 1) | (0/100) |
| Blocks | (Stage II - day 1) | (100/100) |
| Sheep | (Stage II - day 2) | (0/100) |
| Teleportation | (Stage II - day 2) | (0/100) |
| Monotonicity | (Stage III - day 0) | (0/100) |
| Monotonicity 2 | (Stage III - day 0) | (0/100) |
| The Minima Game | (Stage III - day 1) | (100/100) |
| Lamp | (Stage III - day 1) | (0/100) |
| Frog | (Stage III - day 1) | (100/100) |
| Ones | (Stage III - day 2) | (0/100) |
| Bridges | (Stage III - day 2) | (100/100) |
| Pilots | (Stage III - day 2) | (100/100) |
T1 直接黑白染色，如果有个联通块只有一个点就NIE，否则就TAK。
T3 直接暴力枚举长度HASH。用map就行了，不会T的。。总时间nlog^2n。。至于对称，只要把他和他翻转后的乘起来就好辣。。
T5 离线，保存对于每个询问下个需要的数是多少，然后进来一个数就把所有下个需要是这个数的都se掉，这可以用链表存。
T6 裸manacher。
T8 每个询问单独处理。把每个数减去x，他要求的便是最长的一段和>=0的。然后先把前缀和预处理出来，是b[i]，然后他要找i<j，b[i]<=b[j]，然后最大化j-i。这个i一定在从前往后的单调降序列上，这个j一定在从后往前的单调增序列上，然后把这两个序列处理出来就两个指针扫一扫就好辣~
T13 他取的时候一定是把比一个数大的全取光（否则对面可以取你没取的你就亏大了）。所以先排序，然后dp。可以把dp过程优化成两个变量互相转移。
T15 预处理出每个一步会跳到哪里，然后倍增啦~倍增啦~倍增啦~
T17 二分答案，网络流判定。方案可以爆搜。。别怂。。
T18 维护两个单调队列就好了。
（感觉我题解写得认真多了，还有照着bzoj上面AC人数从大到小切真的好么。。）
2022年9月15日 04:27
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