题目大意:给出n对括号,并给出每对括号距离的范围.问能否找到这样一个序列. 题解:好多人都用贪心.这么好的题为什么不搜一发呢? 注意:千万不要在dfs里面更新答案. 代码: #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define N 605 ]; ],g[N][N]; int check(int l,int r,int st) { ; )return f[l]…

Arthur and Brackets 区间dp, dp[ i ][ j ]表示第 i 个括号到第 j 个括号之间的所有括号能不能形成一个合法方案. 然后dp就完事了. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int>…

E. Arthur and Brackets time limit per test 2 seconds memory limit per test 128 megabytes input standard input output standard output Notice that the memory limit is non-standard. Recently Arthur and Sasha have studied correct bracket sequences. Arthu…

题目: E. Arthur and Brackets time limit per test 2 seconds memory limit per test 128 megabytes input standard input output standard output Notice that the memory limit is non-standard. Recently Arthur and Sasha have studied correct bracket sequences. A…

E. Arthur and Brackets time limit per test 2 seconds memory limit per test 128 megabytes input standard input output standard output Notice that the memory limit is non-standard. Recently Arthur and Sasha have studied correct bracket sequences. Arthu…

题目链接:http://codeforces.com/contest/508/problem/E 输入一个n,表示有这么多对括号,然后有n行,每行输入一个区间,第i行的区间表示从前往后第i对括号的左括号跟右括号之间的距离在这个区间的范围内,问是否存在这样的括号序列. 我的做法是比较繁琐,稍微看了下别人的代码,比我的短,应该有更简单 的做法.我的做法是从后往前构造括号序列,每次添加一对括号之前,先把当前的括号序列扫一遍,例如这个括号序列:(())()((()))   ,很显然,现在我要新增一对括号…

大意: n个括号, 位置未知, 给出每对括号左右间距, 求输出一个合法括号序列. 最后一个括号间距一定为1, 从右往左遍历, 每次括号划分越小越好. #include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <s…

n<605设计出n对夸号  给出n个条件每个条件为[l,r] 表示第i对夸号右夸号离左夸号的距离,然后夸号的右夸号出现的顺序必须按照给的顺序 出现, 那么如果存在就输出否则输出impossilbe , 我们知道如果一个夸号在 L位置,那么 另一个夸号就在 L+cnt 这个位置, 那么我们就可以知道在L=1 和L+cnt-1 这之间的夸号只要合法就可以了 #include <iostream> #include <cstdio> #include<algorithm>…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 所给的li,ri是左括号从左到右的顺序给的. (且注意长度是2*n 现在我们先把第一个左括号放在第1个位置. 然后考虑第二个位置. 如果这个位置能放右括号和第一个匹配(位置满足在1+l[i]..1+r[i]之间. 那么我们就在第二个位置放一个右括号就好了. (如果我们作死不放右括号的话,那就只能放左括号了->一定要放一个括号的 那么我们就只能先匹配这一个左括号了,而前一个左括号可能在第3个位置就不能匹配了. 这就会造成错解. 也…

2021-10-14 P2577 [ZJOI2004]午餐 2021-10-13 CF815C Karen and Supermarket(小小紫题,可笑可笑) P6748 『MdOI R3』Fallen Lord(sort(a+1,a+1+n,greater<int>()); 真好用) P4161 [SCOI2009]游戏 P1707 刷题比赛 2021-10-12 CF1573A Countdown P2717 寒假作业 P7868 [COCI2015-2016#2] VUDU P1660…

CF508E Arthur and Brackets 我在赛场上想都没想直接DP \(O(n^3)\)过了 但别人说正解是栈+贪心 讲讲DP \(bool\) \(dp[i][j]\)表示从第i对括号至第j对括号是否在ans中能变成一段连续的区间 转移(\(check(a,b)\)表示\(a=(a || b)\)): 可以第i对括号中间包含第i+1对括号至第j对括号: \(check(dp[i][j],dp[i+1][j])\) 可以由多段相接的连续区间组成(\(i\leq k<j\)): \(…

A. Pasha and Pixels     题意就是给一个n*m的矩阵,k次操作,一开始矩阵全白,一次操作可以染黑一个格子,问第几次操作可以使得矩阵中存在一个2*2的黑色矩阵.直接模拟即可 代码: #include<cstdio> #include<algorithm> #include<map> #include<cstring> #include<vector> #include<cmath> #include<stri…

按下Ctrl + E("编辑")或退出编辑.Brackets将搜索项目下所有CSS文件 Ctrl/Cmd + Alt + P 打开即时预览功能 alt + command + O目录的结构会显示在编辑器的边栏上,点开对应的目录 扩展 Beautify Code Folding Emmet  http://docs.emmet.io/ 直接下载https://github.com/drewhjava/brackets-beautify后,解压放到Brackets 插件目录下(插件目录He…

C. Sereja and Brackets time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja has a bracket sequence s1, s2, ..., sn, or, in other words, a string s of length n, consisting of characters "(…

Brackets是一个开源的前端开发IDE工具,网页设计师和前端开发人员必备的前端开发IDE工具. 它能够使你在开发WEB网站实时预览你的网页,目前版本只适用于Chrome浏览器可以实时预览效果 支持中文,下载下来之后是英文,只需要在Debug中设置语言就可以使用熟悉的原生中文语言. 官网下载地址 http://brackets.io/…

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 27793   Accepted: 7885   Special Judge Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a re…

题意:给括号匹配涂色,红色蓝色或不涂,要求见原题,求方案数 区间DP 用栈先处理匹配 f[i][j][0/1/2][0/1/2]表示i到ji涂色和j涂色的方案数 l和r匹配的话,转移到(l+1,r-1) 不匹配,i的匹配p一定在l和r之间,从p分开转移 听说用记忆化搜索比较快,可以像树形DP那样写记忆化搜索,也可以传统的四个参数那样写 用循环+条件判断,简化状态转移的枚举 注意细节 见代码 #include<iostream> #include<cstdio> #include&l…

是什么? http://brackets.io/ A modern, open source text editor that understands web design. 现代, 开源的文本编辑器, 最懂得web设计. With focused visual tools and preprocessor support, Brackets is a modern text editor that makes it easy to design in the browser. 专注可视化工具…

Coloring Brackets time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Once Petya read a problem about a bracket sequence. He gave it much thought but didn't find a solution. Today you will…

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29502   Accepted: 8402   Special Judge Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a re…

Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [S] are both regular sequences. 3. If A and B are regular sequences, then AB is a regular…

Brackets Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3624   Accepted: 1879 Description We give the following inductive definition of a “regular brackets” sequence: the empty sequence is a regular brackets sequence, if s is a regular…

题目传送门 /* 记忆化搜索(DP+DFS):dp[i][j] 表示第i到第j个字符,最少要加多少个括号 dp[x][x] = 1 一定要加一个括号:dp[x][y] = 0, x > y; 当s[x] 与 s[y] 匹配,则搜索 (x+1, y-1); 否则在x~y-1枚举找到相匹配的括号,更新最小值 */ #include <cstdio> #include <algorithm> #include <cmath> #include <iostream&…

E. Brackets in Implications Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/550/problem/E Description Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is…

C. Arthur and Table Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/557/problem/C Description Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable. In t…

Brackets官网下载: http://brackets.io/ Adobe Brackets是由Adobe主导开发一款主打web开发的编辑器. 是继TextMate,Sublime Text这两个神器后,有望成为Web开发前后端开发神器. Brackets自带强大的插件系统,先已有近百个插件,对nodejs, python, ruby, PHP, Rust等前后端语言都支持非常好.虽然是打着Web开发,从各种插件来看, Brackets的野心是不小的. Brackets的主UI相对清爽,标准…

插件 Brackets Icons  左侧导航的文件图标 FuncDocr  注释工具 QuickDocsJS  js帮助文档 Beautify  格式化代码 Brackets Git  git支持 Emmet  html快速开发 Themes for Brackets  主题集合   主题 Themes for Brackets Dark soda  主题名 14px   字号 Consola  字体…

Brackets Problem Description We give the following inductive definition of a “regular brackets” sequence:● the empty sequence is a regular brackets sequence,● if s is a regular brackets sequence, then (s) are regular brackets sequences, and● if a and…

既求从点(0,0)仅仅能向上或者向右而且不穿越y=x到达点(a,b)有多少总走法... 有公式: C(a+b,min(a,b))-C(a+b,min(a,b)-1)  /// 折纸法证明卡特兰数: http://blog.sina.com.cn/s/blog_6917f47301010cno.html Brackets Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total…

Description We give the following inductive definition of a “regular brackets” sequence: the empty sequence is a regular brackets sequence, if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and if a and b are regul…