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  • poj 2955 Brackets(区间dp)

                                                                                                    Brackets
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 7756   Accepted: 4114

    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

    Source

    以前写过一道类似的题 最优矩阵乘法的那个 也是一道dp 发现这叫区间dp
    dp[i][j]表示【i j]这个区间的最大匹配个数
    我们可以通过求【i,k],[k+1,j]来求这个【i,j]
    所以我们只要枚举这个区间找到最大的那个
    转移方程 dp[i][j]=max(dp[i][k],dp[k+1][j]);
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstdlib>
     4 #include<cctype>
     5 #include<cmath>
     6 #include<cstring>
     7 #include<map>
     8 #include<stack>
     9 #include<set>
    10 #include<vector>
    11 #include<algorithm>
    12 #include<string.h>
    13 typedef long long ll;
    14 typedef unsigned long long LL;
    15 using namespace std;
    16 const int INF=0x3f3f3f3f;
    17 const double eps=0.0000000001;
    18 const int N=1000+10;
    19 char str[N];
    20 int dp[N][N];
    21 int judge(int x,int y){
    22     if(str[x]=='('&&str[y]==')')return 1;
    23     if(str[x]=='['&&str[y]==']')return 1;
    24     return 0;
    25 }
    26 int main(){
    27     while(gets(str)){
    28         if(strcmp(str,"end")==0)break;
    29         memset(dp,0,sizeof(dp));
    30         int len=strlen(str);
    31         for(int t=1;t<len;t++){
    32             for(int i=0;i+t<len;i++){
    33                 int j=i+t;
    34                 if(judge(i,j)){
    35                     if(j-i==1)dp[i][j]=2;
    36                     else dp[i][j]=dp[i+1][j-1]+2;
    37                 }
    38                 for(int k=i;k<j;k++){
    39                     dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
    40                 }
    41             }
    42         }
    43         cout<<dp[0][len-1]<<endl;
    44     }
    45 }
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  • 原文地址:https://www.cnblogs.com/Aa1039510121/p/6814874.html
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