Description ]. Output For each test case, print how many places there are, into which you insert a '(' or ')', can change the sequence to a regular brackets sequence. What's more, you can assume there has at least one such place. Sample Input 4 ) ())…

题目链接:http://poj.org/problem?id=1141 题目大意:给你一串字符串,让你补全括号,要求补得括号最少,并输出补全后的结果. 解题思路: 开始想的是利用相邻子区间,即dp[i+1][j]之类的方法求,像是求回文串的区间DP一样.然后花了3个多小时,GG... 错误数据: (())(]][[)my:6 (()()()[][][][])ans:4 (())([][][][])括号匹配跟回文串不同,并不能通过dp[i+1][j]或者dp[i][j-1]推得dp[i][j],可…

题意:给一段左右小.中括号串,求出这一串中最多有多少匹配的括号. 解法:此问题具有最优子结构,dp[i][j]表示i~j中最多匹配的括号,显然如果i,j是匹配的,那么dp[i][j] = dp[i+1][j-1]+2; 否则我们可以分区间取最值.dp[i][j] = max(dp[i][j],dp[i][k]+dp[k+1][j]); k在i,j之间. 代码: #include <iostream> #include <cstring> #include <algorithm…

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…

题目链接:http://poj.org/problem?id=2955 这题要求求出一段括号序列的最大括号匹配数量 规则如下: 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 regular brackets sequences, th…

冲鸭,去刷题:http://codeforces.com/contest/1153/problem/C C. Serval and Parenthesis Sequence time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Serval soon said goodbye to Japari kindergarten, and b…

Description Sereja has a bracket sequence s1, s2, ..., sn, or, in other words, a string s of length n, consisting of characters "(" and ")". Sereja needs to answer m queries, each of them is described by two integers li, ri(1 ≤ li ≤ ri…

题目描写叙述: 定义合法的括号序列例如以下: 1 空序列是一个合法的序列 2 假设S是合法的序列.则(S)和[S]也是合法的序列 3 假设A和B是合法的序列.则AB也是合法的序列 比如:以下的都是合法的括号序列 (),  [],  (()),  ([]),  ()[],  ()[()] 以下的都是非法的括号序列 (,  [,  ),  )(,  ([)],  ([(] 给定一个由'(',  ')',  '[', 和 ']' 组成的序列,找出以该序列为子序列的最短合法序列. 思路:真是经典的题目.…

题目链接:http://codeforces.com/problemset/problem/5/C 题目大意:给出一串字符串只有'('和')',求出符合括号匹配规则的最大字串长度及该长度的字串出现的次数.解题思路:设dp[i]为到i的最大括号匹配,我们每次遇到一个'('就将其下标存入栈中,每次遇到')'就取出当前栈中里它最近的'('下标即栈顶t.不能直接dp[i]=i-t+1,比如(()()())这样的例子就会出错,应为dp[i]=dp[i-1]+i-t+1. 代码 #include<bits/…

题目链接:http://poj.org/problem?id=2955 题目大意:给你一串字符串,求最大的括号匹配数. 解题思路: 设dp[i][j]是[i,j]的最大括号匹配对数. 则得到状态转移方程: if(str[i]=='('&&str[j]==')'||(str[i]=='['&&str[j]==']')){ dp[i][j]=dp[i+1][j-1]+1; }dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]) ,(i<=k…