题目描述  给定一个1*n的地图,在里面玩2048,每次可以合并相邻两个(数值范围1-40),问最大能合出多少.注意合并后的数值并非加倍而是+1,例如2与2合并后的数值为3. 这道题的思路: 状态: f[i][j] 代表当前 i -> j的最大值 . 初始的f[i][i] = a[i]. 然后枚举点 1 -> n-1 到 n 的区间. 然后中间再枚举断点 k. 方程就是 f[i][j] = max(f[k+1][j]+1,f[i][j]); 约束条件: 当且仅当两个区间所得到的最大值相同的时候…

P3146 [USACO16OPEN]248 题目描述 Bessie likes downloading games to play on her cell phone, even though she doesfind the small touch screen rather cumbersome to use with her large hooves. She is particularly intrigued by the current game she is playing.The…

P3146 [USACO16OPEN]248 题目描述 Bessie likes downloading games to play on her cell phone, even though she doesfind the small touch screen rather cumbersome to use with her large hooves. She is particularly intrigued by the current game she is playing.The…

P3146 [USACO16OPEN]248 题解 第一道自己码出的区间DP快庆祝一哈 2048 每次可以合并任意相邻的两个数字,得到的不是翻倍而是+1 dp[L][R] 区间 L~R 合并结果 然后拆成左区间和右区间,看看他们能不能合并,更新ans 注意如果最后枚举到的总区间 1~n ,那么就要考虑取左右区间最大值了,因为可能左右区间不能合并,那么左右区间最大值就是最终答案 代码 #include<iostream> #include<cstdio> #include<st…

https://www.luogu.org/problemnew/show/P3146 一道区间dp的题,以区间长度为阶段; 但由于要处理相邻的问题,就变得有点麻烦; 最开始想了一个我知道有漏洞的方程 ][j]) f[i][j] = max(f[i][k],f[k + ][j]); ); 可能f[i][k] = f[i][j],但他们可合并的并未相邻; 可以这样 #include <bits/stdc++.h> #define read read() #define up(i,l,r) for…

题目描述 Bessie likes downloading games to play on her cell phone, even though she does find the small touch  screen rather cumbersome to use with her large hooves.She is particularly intrigued by the current g ame she is playing. The game starts with a…

注:两道题目题意是一样的,但是数据范围不同,一个为弱化版,另一个为强化版. P3146传送门(弱化版) 思路: 区间动规,设 f [ i ][ j ] 表示在区间 i ~ j 中获得的最大值,与普通区间动规最大的不同在于:只有左区间的最大值等于右区间的最大值时才能够进行转移. AC代码: #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<cst…

4580: [Usaco2016 Open]248 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 255  Solved: 204[Submit][Status][Discuss] Description Bessie likes downloading games to play on her cell phone, even though she does find the small touch  screen rather cumbers…

题目描述 Bessie likes downloading games to play on her cell phone, even though she doesfind the small touch screen rather cumbersome to use with her large hooves. She is particularly intrigued by the current game she is playing.The game starts with a seq…

记录一些基础的区间 \(\text{DP}\) 题. 0x00 AT_dp_n N - Slimes 最板的区间 \(\text{DP}\) . 记 \(f[i][j]\) 表示合并 \(i\sim j\) 区间的最小代价,初始化为 \(\text{inf}\),\(f[i][i]=0\) (一个数不需要代价). 第一维从小到大枚举区间长度 \(len\),第二维枚举左端点 \(i\),同时可以计算出右端点 \(j=i+len-1\),第三维枚举段点 \(k\),转移方程为 \[f[i][j]=…