CF1266E Spaceship Solitaire

## 题意翻译 Bob在玩一款叫*Spaceship Solitaire*的游戏。游戏的核心是建造一艘宇宙飞船。为了建造，他首先需要积累足够的资源。有$n$种资源，从$1$到$n$标号，Bob需要$a_i$个第$i$种资源来建造飞船。我们称$a_i$为资源$i$的目标。 一回合能且只能生产一个资源。但是，有一些“成就”可以加快生产速度。每个成就可以用一个三元组表示$(s_j, t_j, u_j)$，意思是当Bob拥有的第$s_j$种资源个数达到$t_j$时，他可以获得1个免费的第$u_j$种资源。获得此免费资源有可能使Bob要求获得另一个成就的奖励。 不会有两个成就含有相同的$s_j$和$t_j$，也就是说，达到资源$t_j$个资源$s_j$的奖励最多是一个额外的资源。 对于每个成就，有$0 < t_j < a_{s_j}$ 达到一定资源量的奖励可以是这个资源本身，也就是$s_j = u_j$ 最开始时没有成就的。你需要处理$q$次更新，可能是添加，删除或改变成就。在每次更新后，输出完成游戏（也就是对于每个资源$i \in [1, n]$，收集至少$a_i$个）所需的最小回合数。

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After the first update, the optimal strategy is as follows. First produce $ 2 $ once, which gives a free resource $ 1 $ . Then, produce $ 2 $ twice and $ 1 $ once, for a total of four turns. After the second update, the optimal strategy is to produce $ 2 $ three times — the first two times a single unit of resource $ 1 $ is also granted. After the third update, the game is won as follows. - First produce $ 2 $ once. This gives a free unit of $ 1 $ . This gives additional bonus of resource $ 1 $ . After the first turn, the number of resources is thus $ [2, 1] $ . - Next, produce resource $ 2 $ again, which gives another unit of $ 1 $ . - After this, produce one more unit of $ 2 $ . The final count of resources is $ [3, 3] $ , and three turns are needed to reach this situation. Notice that we have more of resource $ 1 $ than its goal, which is of no use.