1141 Brackets Sequence 括号匹配

 

Brackets Sequence
Time Limit:
1000ms
Memory limit:
10000kB
题目描述
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.

For example, all of the following sequences of characters are regular brackets sequences:

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

And all of the following character sequences are not:

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

Some sequence of characters '(', ')', '[', and ']' is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.
输入
The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.
输出
Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.
样例输入
([(]
样例输出
()[()]

//

 
#include<iostream>
#include<cstring>
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))
    {
        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;
                 }
             }
         }
        out(0,n-1);cout<<endl;
        //cout<<dp[0][n-1]<<endl;
     }
     return 0;
 }