joj 2262: Brackets 最多有多少括号匹配

最新推荐文章于 2020-05-17 17:24:15 发布

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本文探讨了如何找出给定括号序列中最长的有效括号子序列的长度，并提供了一个具体的实现方案。通过动态规划的方法，算法能够高效地解决这一问题。

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 a1a2 . . . an, 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 < . . . < i_m ≤ n, a_i₁ a_i₂ . . . a_{i_m} is a regular brackets sequence. For example, 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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
int dp[115][115],path[115][115];
char s[115];
void out(int i,int j)
{
    if(i>j) return ;
    if(i==j)
    {
        if(s[i]=='['||s[i]==']') cout<<"[]";
        else cout<<"()";
        return ;
    }
    if(path[i][j]==-1)
    {
        cout<<s[i];
        out(i+1,j-1);
        cout<<s[j];
    }
    else
    {
        out(i,path[i][j]);
        out(path[i][j]+1,j);
    }
}
int main()
{
    while(gets(s))
    {
        if(s[0]=='e') break;
        memset(path,0,sizeof(path));
         int n=strlen(s);
         if(n==0) {cout<<endl;continue;}
         for(int i=0;i<=n;i++) dp[i][i]=1,dp[i][i-1]=0;
         for(int p=1;p<n;p++)
         {
             for(int i=0;i+p<n;i++)
             {
                 int j=i+p;
                 dp[i][j]=(1<<31)-1;
                 if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
                 {
                     if(dp[i][j]>dp[i+1][j-1]) dp[i][j]=dp[i+1][j-1],path[i][j]=-1;
                 }
                /* if(s[i]=='('||s[i]=='[')
                 {
                     if(dp[i][j]>dp[i+1][j]+1) dp[i][j]=dp[i+1][j]+1,path[i][j]=i;
                 }
                 if(s[j]==')'||s[j]==']')
                 {
                     if(dp[i][j]>dp[i][j-1]+1) dp[i][j]=dp[i][j-1]+1,path[i][j]=j-1;
                 }*/ //inclued by the next
                 for(int k=i;k<j;k++)
                 {
                     if(dp[i][j]>dp[i][k]+dp[k+1][j]) dp[i][j]=dp[i][k]+dp[k+1][j],path[i][j]=k;
                 }
             }
         }
        cout<<n-dp[0][n-1]<<endl;
     }
     return 0;
}