1141Brackets Sequence结题报告

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23638		Accepted: 6665		Special Judge

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

()[()]

题目分析：
　　显然题目是要求我们为一个输入序列添加最少的“[”或者“(”等符号来使之可以配对。要配对的办法肯定有很多种，我们检测到每一个括号都给他配对，那么最终多出了一倍的符号，这样肯定是配对的但是不符合最少的。因此我们采用一个区间DP的方法。pos[i][j]表示第i个和第j个字符之间的需要隔开的位置，若是匹配的则为-1，dp[i][j]表示第i个和第j个字符之间需要添加的字符数。具体见如下Java代码注释。

import java.util.*;
public class BracketsSequence1141 {
    private static char s[]=new char[105];//接受控制台的输入。
    private int len;//输入字符串的长度
    private int dp[][]=new int[105][105];//dp[i][j]，从i到j需要插入的最小字符数
    private int pos[][]=new int[105][105];//pos[i][j],字符i与j是否匹配，若是则为-1，若不是则记录隔开的位置
    private void print(int i,int j){
        //输出结果
        if(i>j)
            return;
        if(i==j){
            if(s[i]=='('||s[i]==')')
                System.out.print("()");
            else
                System.out.print("[]");
        }
        else{
            if(pos[i][j]==-1){//如果字符是匹配的
                System.out.print(s[i]);
                print(i+1,j-1);
                System.out.print(s[j]);
            }
            else{//字符不匹配
                print(i,pos[i][j]);
                print(pos[i][j]+1,j);
            }
        }
    };
    private void getDP(){
        int inf=0x7fffffff;
        int i,j,k,mid;
        int len=s.length;
        for(i=0;i<len;i++){
            for(j=0;j<len;j++){
                dp[i][j]=0;
            }
            dp[i][i]=1;
        }
        for(k=1;k<len;k++){
            for(i=0;i+k<len;i++){
                j=i+k;//表示不断地分隔，比较不同字符之间是否匹配
                dp[i][j]=inf;//表示目前不可匹配
                if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')){
                    //表示字符配对
                    dp[i][j]=dp[i+1][j-1];
                    pos[i][j]=-1;
                }
                //若字符不配对或者其他，表示寻找更少的配对数
                for(mid=i;mid<j;mid++){
                    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;
                    }
                }
            }
        }
    }
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        //String temp=null;
        BracketsSequence1141 A=new BracketsSequence1141();
        s=sc.nextLine().toCharArray();
        if(s.length==0){
            System.out.println();
        }
        else{
            A.getDP();
            A.print(0,s.length-1);
        }    
    }    
}