面试题库

14 天用 Agent 搭一条 AI 短剧生产线

D3 角色一致性：定妆图、参考图与风格锁定，让同一个人每一镜都还是他

• 生成类资产要做复用，缓存键你会怎么设计，才能既省钱又不会串戏？How would you design the cache key for reusing generated assets so that you save money without serving the wrong asset?

  国内高频海外高频深入#caching#cost#image-generation

  分析过程 · 先想清楚再作答

  1. 这题考的是缓存的两类错误，而且两类的代价完全不对称。少命中只是多花钱，错命中会把上一集的道具塞进这一集——前者可量化，后者是内容事故。
  2. 推导链只有一句：**键必须由所有会改变产物的输入算出来，一项不多一项不少。** 多算了不该算的（比如输出路径），改一次目录结构缓存全部失效，白花一遍钱；少算了该算的（比如提示词），换了描述还命中老图，就是串戏。
  3. 落到这个场景，参与哈希的是：资产类别、归属对象、变体名、完整提示词、参考图标识、随机种子。用 sha1 之类取个短摘要当 id，元数据里再把这几项原样存一份，出问题能照着复现。
  4. 然后主动把边界说清楚，这是加分项：模型 id 与版本要不要进键？要。风格模板改了怎么办？它是提示词的一部分，进键之后天然全部失效——所以模板要谨慎改，或者给它一个版本号，让你能决定失效的范围。
  5. 生产视角还有一条：失败的生成不要写进缓存，否则你会稳定复用一张被审核拦下的空结果。命中缓存的那条路径也要记台账并标成命中，不然你算不出缓存到底省了多少钱。
  6. 可以预期的追问：缓存要不要过期？答案是内容型资产通常不设时间过期，而是靠版本号显式失效；时间过期会在你毫无预期的时候让一整集重新生成一遍。

  How to reason about it · think before answering

  1. This question is about two kinds of cache error with wildly asymmetric cost. A miss only costs money; a wrong hit puts last episode's prop into this one. The first is a number, the second is a content incident.
  2. The derivation is one sentence: the key must be computed from every input that changes the artifact, and nothing else. Include something irrelevant, like the output path, and one directory refactor invalidates everything and you pay again; omit something relevant, like the prompt, and a changed description silently serves the old image.
  3. Concretely, hash the asset kind, the owning entity id, the variant name, the full prompt, the reference image identity and the seed. Take a short digest as the id, and store those fields verbatim in the metadata so any artifact can be reproduced.
  4. Then name the boundaries yourself: does the model id and version belong in the key? Yes. What if the style template changes? It is part of the prompt, so it invalidates everything by construction — which is why templates should carry a version number, letting you choose the blast radius.
  5. One more production note: never cache failed generations, or you will faithfully reuse an empty result that safety review rejected. Cache hits also belong in the cost ledger, flagged as hits, otherwise you cannot report how much caching saved.
  6. Expect the follow-up: should the cache expire? Content assets usually should not expire on time; invalidate explicitly by version instead, because a time-based expiry regenerates a whole episode at the least convenient moment.

  答题要点

  • 键由所有会改变产物的输入算出：资产类别、归属对象、变体名、完整提示词、参考图标识、随机种子，再加模型 id 与版本
  • 不要把输出路径或文件名放进键，改目录结构会让缓存整体失效，白付一遍钱
  • 少算提示词这类输入会导致错命中，那是内容事故，代价远高于少命中
  • 元数据里原样保存参与哈希的各项，出问题能复现；失败的生成不写缓存
  • 命中缓存也要记台账并标成命中，否则算不出缓存省了多少；失效靠显式版本号而不是时间过期

  Key points

  • Derive the key from everything that changes the artifact: asset kind, owner id, variant, full prompt, reference image identity, seed, plus model id and version
  • Keep output paths and filenames out of the key, or one directory refactor invalidates the whole cache and you pay twice
  • Omitting inputs like the prompt causes wrong hits, which are content incidents and far costlier than misses
  • Store the hashed fields verbatim in metadata so any artifact is reproducible, and never cache failed generations
  • Record cache hits in the cost ledger flagged as hits, and invalidate explicitly by version rather than by time

D4 从分镜到镜头：图生视频、异步任务轮询与失败重试

• 生成类接口返回失败，你怎么判断该不该重试？重试几次之后该做什么？When a generation API returns a failure, how do you decide whether to retry, and what happens after the retries run out?

  国内高频海外高频深入#error-handling#retry#cost

  分析过程 · 先想清楚再作答

  1. 这题的题眼是「判断」。按状态码首位数字一刀切是最常见的错误答案，因为生成类接口的业务错误码往往和 HTTP 状态码不在一个层面上——很多厂商的失败是 HTTP 200 加一个响应体里的业务码。
  2. 给一条可复用的判据，比背错误码表有用：问三个问题——等一等会不会好、改输入会不会好、还是必须叫人来。三个问题对应三种处置：退避重试、修请求、立刻告警。
  3. 落到具体：限流和服务端故障属于第一类，程序自己扛；参数无效与内容审核属于第二类，重试一万次都是同一个错，而且会挤占限流额度让真正该重试的排不上号；鉴权失败与余额不足属于第三类，重试只会延迟告警。
  4. 然后单独处理超时，这是最能体现经验的一条：超时不是失败，是状态未知，对方队列里那个任务可能还在跑甚至已经成了。所以超时之后不能直接重提，要先按幂等键查一遍已有产物。
  5. 重试用尽之后要做三件事，缺一不可：把这一条标成失败并记下最后一次的错误码与请求参数、继续跑批次里剩下的任务不要中断、把失败清单汇总成一次可读的告警而不是每条发一次。
  6. 可以预期的追问：重试次数怎么定？按单价定。单价越高，允许的重试次数越少，而且高单价的失败更应该先送人复核再决定要不要重做。

  How to reason about it · think before answering

  1. The hinge is 'decide'. Bucketing by the leading digit of the HTTP status is the classic wrong answer, because generation APIs often return HTTP 200 with a business error code in the body.
  2. Give a reusable test instead of reciting a code table: ask three questions — will waiting help, will changing the input help, or does a human have to step in? They map onto three dispositions: back off and retry, fix the request, alert immediately.
  3. Concretely: rate limits and server errors are the first bucket and the program handles them; invalid parameters and content moderation are the second, where retrying repeats the same error and burns rate-limit budget that genuinely retryable tasks needed; auth failure and insufficient balance are the third, where retrying only delays the alert.
  4. Handle timeout separately — this is the line that signals experience. A timeout is unknown, not failed: the job may still be running, or may have finished. So never resubmit blindly; look up the idempotency key for an existing artifact first.
  5. When retries are exhausted, do three things: mark the item failed with the last error code and the exact request parameters, keep processing the rest of the batch instead of aborting it, and aggregate the failures into one readable alert rather than one per item.
  6. Expect the follow-up: how many retries? Scale it by unit price. The more expensive the call, the fewer automatic retries, and expensive failures should go to a human for review before being redone.

  答题要点

  • 不要按状态码首位一刀切，生成类接口的业务错误码常常藏在 HTTP 200 的响应体里
  • 判据是三个问题：等一等会不会好、改输入会不会好、还是必须叫人来，分别对应退避重试、修请求、立刻告警
  • 限流与服务端故障可重试；参数无效与内容审核重试无用且会挤占限流额度；鉴权失败与余额不足必须告警
  • 超时是状态未知不是失败，重试前先按幂等键查一遍已有产物，否则会重复计费
  • 重试用尽后：标记失败并留下错误码与请求参数、不中断整批、把失败汇总成一次可读告警；重试次数按单价定

  Key points

  • Do not bucket by the leading HTTP digit; generation APIs often hide the business error code inside an HTTP 200 body
  • Use three questions — will waiting help, will changing the input help, or is a human required — mapping to back off, fix the request, alert
  • Rate limits and server errors are retryable; invalid parameters and moderation blocks are not and waste rate-limit budget; auth and balance failures need an alert
  • A timeout is unknown rather than failed: check the idempotency key for an existing artifact before resubmitting, or you pay twice
  • When retries run out, mark the item failed with its error code and request parameters, keep the batch running, and aggregate failures into one alert; scale retry counts by unit price

D10 审片室：能预览、能改词、能重生成单镜的人机协作后台

• 一条自动化流水线要插入人工审核，你会把卡点放在哪几步？为什么？Where would you place human review checkpoints in an automated pipeline, and why there?

  国内高频海外高频基础#human-in-the-loop#pipeline-design#cost

  分析过程 · 先想清楚再作答

  1. 这题在考你有没有成本意识。答「每一步都让人看一眼」是没做过工程的回答——人是最贵的资源，卡点多了流水线就退化成手工作坊。
  2. 给一条可复用的判据：**卡点放在「下游最贵的那一步」之前**。判断某个位置该不该设卡，只问一句「如果这里错了，往后要白花多少钱」。
  3. 按这条判据落到生成式流水线上，会得到三个位置：剧本定稿之后（此时零成本，却决定了后面所有素材的方向）、首帧出来之后视频生成之前（首帧是最便宜的一档，视频是最贵的一档，同一个镜头差出一到两个数量级）、成片合成之后发布之前（这一道拦的不是质量而是合规风险）。
  4. 补一条生产视角：卡点不等于阻塞。第一和第二道可以做成「默认放行、超时自动继续」，只有第三道必须硬卡——合规问题不能靠超时放行。
  5. 结论里要点出一个反直觉的事实：最容易被跳过的恰恰是第一道，因为这时候还没有画面，看起来没什么可审的；但它是唯一一道改起来零成本的闸门。
  6. 可预期的追问是「人来不及审怎么办」。答案是分级：机器先打分，只把低分的推给人，人的时间花在机器拿不准的那部分上。

  How to reason about it · think before answering

  1. This one tests cost awareness. Saying a human should look at every step marks someone who has not run this in production: humans are the expensive resource, and too many gates turn a pipeline back into handwork.
  2. Offer a reusable rule: put the gate immediately before the most expensive downstream step. To decide whether a position deserves a gate, ask how much money is wasted if something is wrong here.
  3. Applied to a generative pipeline that yields three positions: after the script is locked (free to change, yet it steers every asset that follows), after the first frame but before video generation (the frame is the cheapest step and the clip is the most expensive, one to two orders of magnitude apart), and after the final cut but before publishing (this one gates risk, not quality).
  4. Add the production view: a checkpoint is not necessarily blocking. The first two can auto-continue on timeout; only the compliance gate must hard-block, because you cannot let a legal check pass by timing out.
  5. State the counterintuitive part: the first gate is the one people skip, because there are no visuals yet and it looks like there is nothing to review, while it is the only gate where changes cost nothing.
  6. Expected follow-up: what if reviewers cannot keep up. Tier it. Machines score everything, humans only see the low scores, and human attention goes where the machine is unsure.

  答题要点

  • 判据是「卡点放在下游最贵的那一步之前」，问的是这里错了往后白花多少钱。
  • 三个位置：剧本定稿后、首帧出来后视频生成前、成片合成后发布前。
  • 首帧那一道性价比最高：首帧是最便宜的一档，视频是最贵的一档，同一个镜头差出一到两个数量级。
  • 前两道可以默认放行加超时继续，只有合规那一道必须硬卡。
  • 人力不够就分级：机器先打分，人只看低分的那些。

  Key points

  • Rule: place the gate right before the most expensive downstream step, judged by wasted spend if this step is wrong.
  • Three positions: after script lock, after first frame and before video, after final cut and before publish.
  • The first-frame gate pays best: the frame is the cheapest step and the clip the most expensive, one to two orders of magnitude apart.
  • The first two gates can auto-continue on timeout; only the compliance gate hard-blocks.
  • When reviewers are the bottleneck, tier it: machines score everything, humans only see low scores.
• 用户改了中间一步的输入，怎么算出哪些下游需要重做？A user edits an intermediate input. How do you compute which downstream steps must rerun?

  国内高频海外高频进阶#dag#incremental-recompute#cost

  分析过程 · 先想清楚再作答

  1. 这题的区分度在方向和收尾两处，很多人只答出中间那段「沿依赖图传播」，前后都丢了。
  2. 方向：从被改的节点**沿着「谁依赖我」正向传播**，不是往上游找依赖。写反的后果很隐蔽——上游会被一起重跑，结果是对的，钱多花了一倍，测试也发现不了。
  3. 落到实现：把种子节点放进集合，反复扫一遍图，只要某个节点的依赖里有一个已经在集合里就把它也加进来，跑到不动点为止；最后按拓扑序返回，调用方顺着数组跑就不会先跑下游后跑上游。
  4. 收尾这一步最容易漏：**没受影响的节点，产物要从上一版复制过来，不是重新生成**。半径算得再准，少了复制这一步就一分钱没省。
  5. 然后是怎么验证。不要比文件哈希——同样的输入很可能生成逐字节相同的结果，哈希相同证明不了没重跑。要数**接口调用次数**，这才是硬证据，而且在离线与真实两种模式下都成立。
  6. 可预期的追问是「输入没变但你想重跑怎么办」。留一个强制重跑的开关，并且把它和自动判定分开记账，否则你会分不清一次重跑是系统判的还是人手动点的。

  How to reason about it · think before answering

  1. The signal lives at the two ends. Most candidates produce the middle part, propagation over a dependency graph, and drop both the direction and the finish.
  2. Direction: propagate forward along who-depends-on-me from the edited node, not backward to its dependencies. Getting it backward is insidious, because upstream nodes rerun, the output is still correct, the bill doubles, and no test catches it.
  3. Implementation: seed a set, sweep the graph repeatedly adding any node with a dependency already in the set until it stops growing, then return in topological order so the caller can just walk the array.
  4. The finish is what people forget: unaffected nodes must have their artifacts copied from the previous version, not regenerated. A perfect radius saves nothing without that copy.
  5. Then verification. Do not compare file hashes, because identical inputs often produce byte-identical output and a matching hash proves nothing. Count API calls instead; that evidence holds both offline and against a real vendor.
  6. Expected follow-up: what about forcing a rerun when nothing changed. Keep an explicit force flag and account for it separately, or you lose the ability to tell system-decided reruns from human-triggered ones.

  答题要点

  • 从被改的节点沿着「谁依赖我」正向传播，不是反向找依赖。
  • 扫图到不动点，结果按拓扑序返回，保证执行顺序不会颠倒。
  • 没受影响的节点要从上一版复制产物，否则半径算得再准也没省钱。
  • 验证要数接口调用次数，不要比文件哈希——同样的输入可能产出逐字节相同的结果。
  • 另留一个强制重跑开关，并与自动判定分开记账。

  Key points

  • Propagate forward along who-depends-on-me from the edited node, never backward.
  • Sweep to a fixed point and return in topological order so execution never runs downstream first.
  • Copy artifacts for unaffected nodes from the previous version, or the computed radius saves nothing.
  • Verify by counting API calls, not by comparing file hashes, since identical inputs can produce byte-identical output.
  • Keep a separate force-rerun switch and account for it apart from automatic decisions.