POJ 1141 Brackets Sequence（区间dp + dfs）

题目链接http://poj.org/problem?id=1141

                                                                                           Brackets Sequence

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

()[()]

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int maxn=220;const int inf=0x7fffffff;int pos[maxn][maxn];//从i到j在哪里分开int dp[maxn][maxn];//从i到j至少添几个符号char str[maxn];int len;void print(int i,int j){  if(i>j)    return ;//递归出口  if(i==j)  {    if(str[i]=='('||str[i]==')')      cout<<"()";    else      cout<<"[]";  }  else if(pos[i][j]==-1)//两边是对称的  {    cout<<str[i];    print(i+1,j-1);    cout<<str[j];  }  else//可分割  {    print(i,pos[i][j]);    print(pos[i][j]+1,j);  }}int main(){  cin>>str;  len=strlen(str);  memset(dp,0,sizeof(dp));  for(int i=0;i<len;i++)    dp[i][i]=1;  for(int k=1;k<len;k++)//长度    for(int i=0;i+k<len;i++)//起点    {      int j=i+k;      dp[i][j]=inf;      if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']'))      {        dp[i][j]=dp[i+1][j-1];        pos[i][j]=-1;//暂时让它等于-1      }      for(int mid=i;mid<j;mid++)//这个必须要执行的。      {        if(dp[i][j]>(dp[i][mid]+dp[mid+1][j]))        {          dp[i][j]=dp[i][mid]+dp[mid+1][j];          pos[i][j]=mid;        }      }    }  print(0,len-1);  cout<<endl;  return 0;}