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 区间dp,这次设计的状态和一般的有一定的差异. 这次我们定义$dp[i][j]$表示$[i,j]$的可以合并出来最大取值,而不是合并区间$[i,j]$的最大取值. 同样的我们枚举区间长度,枚举左端点,求出右端点. 枚举$i$到$j$之间的每一个分割点,判断两点之间是否可以合并,取价值更高的答案. $$dp[i][j]=max(dp[i][j],dp[i][k]+1) [dp[i][k]==dp[i][k+1]…

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…

P3143 [USACO16OPEN]钻石收藏家Diamond Collector 题目描述 Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare time! She has collected \(N\) diamonds (\(N \leq 50,000\) of varying sizes, and she wants to arrange so…

P3143 [USACO16OPEN]钻石收藏家Diamond Collector 题目描述 Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare time! She has collected NN diamonds (N \leq 50,000N≤50,000) of varying sizes, and she wants to arrange…

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…

题目描述 Farmer John and his cows are planning to leave town for a long vacation, and so FJ wants to temporarily close down his farm to save money in the meantime. The farm consists of NN barns connected with MM bidirectional paths between some pairs of…

传送门 题目大意: n个谷仓 ,每次关闭一个谷仓,问剩下没被关闭的谷仓是 否联通. 题解:并查集+倒序处理 代码: #include<iostream> #include<cstdio> #include<cstring> #define N 3030 using namespace std; int n,m,sumedge,cnt; int head[N],fa[N],q[N],ans[N],exit[N]; struct Edge{ int x,y,nxt; Edg…