题目描述 Farmer John is arranging his NN cows in a line to take a photo (1 \leq N \leq 501≤N≤50). The height of the iith cow in sequence is a(i)a(i), and Farmer John thinks it would make for an aesthetically pleasing photo if the cow lineup has a large i…

嘟嘟嘟 这题刚开始是什么思路也没有,关键是不知道怎么解决序列反转的问题. 然后我就想到如果暴力反转一个序列的话,实际上就是不断交换数组中的两个数ai和aj,同时要满足交换的数不能交叉. 然后又看了一眼(岂止一眼)题解,因为ai <= 50,所以令dp[i][j][L][R]表示区间[i, j],min(ak) >= L, max(ak) <= R时,反转一次的最长不下降子序列. 显然是一个区间dp,那么[i, j]可以从[i + 1, j],[i, j - 1]或是[i + 1, j -…

题目大意:给你一个序列,你可以翻转任意一段子序列一次,求最长不下降子序列长度 tips:子序列可以不连续,但不能破坏在原序列中的顺序 观察数据范围,n<=50,很小,考虑dfs *dfs来跑区间dp可以巧妙的解决区间两端元素的置换问题 记忆化搜索,f[i][j][l][r] 代表对于区间[i,j],构成最长不下降子序列的元素值域在[l,r]时,最长不下降子序列的长度 注意特判端点 #include <bits/stdc++.h> #define N 55 #define inf 0x3f…

4758: [Usaco2017 Jan]Subsequence Reversal Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 76  Solved: 52[Submit][Status][Discuss] Description Farmer John is arranging his NN cows in a line to take a photo (1≤N≤50). The height of the iith co w in sequ…

因为之前忘做了,赶紧补上. T1.Promotion Counting 题目大意:给定一个以1为根的N个节点的树(N<=100,000),每个节点有一个权值,对于每个节点求出权值比它大的子孙的个数. 思路:肯定先要求出dfs序,首先无脑想到主席树,后来发现只要按权值从大到小处理就不用那么麻烦了. #include<cstdio> #include<algorithm> using namespace std; <<],*S=B,C;int X; inline in…

区间\(dp\)提升复习 不得不说这波题真的不简单... 技巧总结: 1.有时候转移可以利用背包累和 2.如果遇到类似区间添加限制的题可以直接把限制扔在区间上,每次只考虑\([l,r]\)被\([i,j]\)完全包含的情况 [BZOJ4897] [Thu Summer Camp2016] 成绩单 典型的利用背包转移,\(dp[i][j]\)表示处理完这段\([i,j]\)区间的答案 转移时一段区间可以被先处理掉或者直接从已经处理完的\(dp[i][j]\)中取过来,这个可以用一个背包维护 con…

A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. For example, these are arithmetic sequences: 1, 3, 5, 7, 9 7, 7, 7, 7 3, -1, -5, -9 The follo…

Given a string s and a string t, check if s is subsequence of t. You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100). A subsequence of…

A sequence of numbers is called a wiggle sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two…

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…