标签：pairs R1 R2 Dirt Deeds R3 Cheap max type
First I thought about modelling this problem as a directed graph, where between each pair of nodes, an edge represents a valid < > or > < transition. Then do a dfs with dp to compute the longest valid path. But there can be as many as O(N^2) edges.
By drawing some examples, I made the following observations:
1. there are 2 types of integer pairs, either (a > b) or (a < b).
2. we will never mix these 2 types together in any valid path. For (a < b), the next pair must be (c, d) such that b > c, c < d. So we separate these 2 types and consider them separately.
3. For each type, greedy works: For type a < b, pick max(b) first; for type a > b, pick min(b) first. Doing this allows us to use all pairs of the same type.
Proof: For type a < b, say we have (L1, R1), (L2, R2), (L3, R3), without the loss of generality, let's assume R1 > R2 > R3. So R1 is the max, R2 is the second max and R3 is the third max and we have R1 > R2 > L2, R2 > R3 > L3, we can connect all 3 pairs in this order.
标签：pairs,R1,R2,Dirt,Deeds,R3,Cheap,max,type 来源： https://www.cnblogs.com/lz87/p/15824698.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。