正规括号序列定义为: 空序列是正规括号序列 如果S是正规括号序列,那么[S]和(S)也是正规括号序列 如果A和B都是正规括号序列,则AB也是正规括号序列 输入一个括号序列,添加尽量少的括号使之成为正规括号序列,并输出最优方案,多解的话输出任意一个即可. 设d(i, j)表示字符串s[i]~s[j]至少添加的括号的数量,则转移如下: S形如[S']或(S'),则转移到d(i+1, j-1) 如果S至少有两个字符,将其分为AB,转移到min{d(i, j), d(A) + d(B)} 不管是否满足第…

[Link]: [Description] 括号序列由这样的规则生成: 1.空字符是一个括号序列; 2.在括号序列两端加上一对括号也是括号序列; 如(s),[s]; 3.两个括号序列A和B,连在一起,也是一个括号序列,即AB也是括号序列 给你一个只包含"()[]"这4种字符的字符串; 让你添加最少的括号,使得所成的序列是一个括号序列; [Solution] 根据括号序列的生成规则; 设dp[l][r]是,l..r这一段变为合法的括号序列需要添加的括号个数; dp[i][i] = 1;…

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

[poj P1141] Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K   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 regular sequence, then (S) and […

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…

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…

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&…

UVA 1546 - Complete the sequence! 题目链接 题意:给定多项式前s项,求出后c项,要求尽量小 思路:利用差分法,对原序列求s - 1次差分,就能够发现规律,然后对于每多一项,就逆推回去就可以 代码: #include <stdio.h> #include <string.h> const int N = 205; int t, s, c, a[N][N]; int main() { scanf("%d", &t); whi…

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 sequence. F…