括号序列（Poj1141）

Poj1141

题目描述：

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

(,  [,  ),  )(,  ([)],  ([(] 

给定一个由'(',  ')',  '[', 和 ']' 组成的序列，找出以该序列为子序列的最短合法序列。

Brackets Sequence

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 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.

Input

The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.

Output

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.

Sample Input

([(]

Sample Output

()[()]

分析：先用动态规划处理一个f[i][j]数组表示把i-j之间的括号处理成合法序列的最
小步数，处理完成之后，再用深搜把序列打出来
代码：
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define M 210
using namespace std;
char s[M];
int f[M][M];
bool check(char x,char y)
{
    if(x=='('&&y==')') return true;
    if(x=='['&&y==']') return true;
    return false;
}
void work(int l,int r)
{
    if(l>r) return ;
    if(l==r)
    {
        if(s[l]=='('||s[r]==')') printf("()");
        if(s[l]=='['||s[r]==']') printf("[]");
        return ;
    }
    int tot=f[l][r];
    if(tot==f[l+1][r-1]&&check(s[l],s[r]))
    {
        printf("%c",s[l]);
        work(l+1,r-1);
        printf("%c",s[r]);
        return;
    }
    for(int k=l;k<r;k++)
      if(tot==f[l][k]+f[k+1][r])
      {
          work(l,k);work(k+1,r);return ;
      }
}
int main()
{
    freopen("jh.in","r",stdin);
    scanf("%s",s+1);
    int n=strlen(s+1);
    for(int w=1;w<=n;w++)f[w][w]=1;
    for(int l=1;l<n;l++)
      for(int i=1;i<=n-l;++i)
        {
            int j=l+i;
            f[i][j]=9999999;
            if(check(s[i],s[j]))
              f[i][j]=f[i+1][j-1];
            for(int k=i;k<=j-1;k++)
              if(f[i][k]+f[k+1][j]<f[i][j])
                  f[i][j]=f[i][k]+f[k+1][j];
        }
    work(1,n);
    printf("
");
    return 0;
}