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2020-2021:teams:alchemist:weekly_digest_10 [2020/08/14 17:49]
hardict [龙鹏宇 Hardict]
2020-2021:teams:alchemist:weekly_digest_10 [2020/08/14 22:11] (当前版本)
maxdumbledore
行 89: 行 89:
   - $dp[i][j][0/​1]$代表前$i$个线段中钦定的$X$为$j$,是/​否有一个线段的右端点为$X$   - $dp[i][j][0/​1]$代表前$i$个线段中钦定的$X$为$j$,是/​否有一个线段的右端点为$X$
   - $dp[i][j][0/​1]=dp[i-1][j][0/​1]\ (j<L[i] \lor j>R[i])$   - $dp[i][j][0/​1]=dp[i-1][j][0/​1]\ (j<L[i] \lor j>R[i])$
-  - $dp[i][j][0/1]=dp[i-1][j][0/1]*2+(j-L[i]^2)\ (L[i]\le j<​R[i])$ +  - $dp[i][j][0]=dp[i-1][j][0]\times ​2+(j-L[i])^2,​dp[i][j][1]=dp[i-1][j][1]\times 2\ (L[i]\le j<​R[i])$ 
-  - $dp[i][R[i]][0]=dp[i-1][R[i]][0]+  - $dp[i][R[i]][0]=dp[i-1][R[i]][0],dp[i][R[i]][1]=dp[i-1][R[i]][0]+dp[i-1][R[i]][1]\times ​2+(R[i]-L[i])^2$
-  - $dp[i][R[i]][1]=dp[i-1][R[i]][0]+dp[i-1][R[i]][1]*2+(R[i]-L[i])^2$+
  
 最后使用线段树或其他数据结构就可以将该$dp$优化到$O(n\log n)$ 最后使用线段树或其他数据结构就可以将该$dp$优化到$O(n\log n)$
2020-2021/teams/alchemist/weekly_digest_10.1597398553.txt.gz · 最后更改: 2020/08/14 17:49 由 hardict