A bottom-up DP. To be honest, it is not easy to relate DP to this problem. Maybe, all "most"\"least" problems can be solved using DP.. Reference: http://blog.sina.com.cn/s/blog_8e6023de01014ptz.html There's an important details to AC:…

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29520   Accepted: 8406   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…

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35049   Accepted: 10139   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 r…

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…

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…

题目链接:http://poj.org/problem?id=1141 题解:求已知子串最短的括号完备的全序列 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; #define ll long long ; const int INF=0x3f3f3f3f; ][]; ][]; ]; int Find(int x…

题目:http://poj.org/problem?id=1141 转载:http://blog.csdn.net/lijiecsu/article/details/7589877 定义合法的括号序列如下: 1 空序列是一个合法的序列 2 如果S是合法的序列,则(S)和[S]也是合法的序列 3 如果A和B是合法的序列,则AB也是合法的序列 例如:下面的都是合法的括号序列 (),  [],  (()),  ([]),  ()[],  ()[()] 下面的都是非法的括号序列 (,  [,  ),  …

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…

题目链接 很早 很早之前就看过的一题,今天终于A了.状态转移,还算好想,输出路径有些麻烦,搞了一个标记数组的,感觉不大对,一直wa,看到别人有写直接输出的..二了,直接输出就过了.. #include <cstdio> #include <cstring> #include <iostream> using namespace std; ][]; ]; ]; ]; int dfs(int L,int R) { int i,minz; if(dp[L][R]) retur…