zoukankan      html  css  js  c++  java
  • Brackets

    Brackets
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 3871   Accepted: 2028

    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 regular brackets sequences, then ab is a regular brackets sequence.
    • no other sequence is a regular brackets sequence

    For instance, all of the following character sequences are regular brackets sequences:

    (), [], (()), ()[], ()[()]

    while the following character sequences are not:

    (, ], )(, ([)], ([(]

    Given a brackets sequence of characters a1a2an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < imn, ai1ai2aim is a regular brackets sequence.

    Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

    Input

    The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

    Output

    For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

    Sample Input

    ((()))
    ()()()
    ([]])
    )[)(
    ([][][)
    end

    Sample Output

    6
    6
    4
    0
    6
     1 #include<stdio.h>
     2 #include<string.h>
     3 #define max(x,y) x>y?x:y
     4 int main(){char sequence[110];
     5     int dp[105][105],t;
     6     while(scanf("%s",sequence),strcmp(sequence,"end")){
     7         memset(dp,0,sizeof(dp));
     8         t=strlen(sequence);
     9         for(int i=t-2;i>=0;i--){
    10             for(int j=i+1;j<t;j++){dp[i][j]=dp[i+1][j];
    11                 for(int k=i+1;k<=j;k++){
    12                     if(sequence[i]=='('&&sequence[k]==')'||sequence[i]=='['&&sequence[k]==']'){
    13                         dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k][j]+2);
    14                     //    printf("%d %d %c %c %d
    ",i,k,sequence[i],sequence[k],dp[i][k]);
    15                     }
    16                 }
    17             }
    18         }
    19         printf("%d
    ",dp[0][t-1]);
    20     }
    21     return 0;
    22 }
  • 相关阅读:
    ZOJ 2671 Cryptography(线段树+求区间矩阵乘积)
    HDU 4662 MU Puzzle(找规律)
    Codeforces 392 C Unfair Poll(模拟)
    UVA 11134 Fabled Rooks(传说中的车)(贪心)
    UVA 11212 Editing a Book(IDA*算法+状态空间搜索)
    用户的昵称【哈希】
    【洛谷P2375】动物园【KMP】
    【洛谷P2375】动物园【KMP】
    【洛谷P2375】动物园【KMP】
    【洛谷P1886】滑动窗口【单调队列】
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4534995.html
Copyright © 2011-2022 走看看