首页 > 代码库 > 1141Brackets Sequence结题报告
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.
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。
1 import java.util.*; 2 public class BracketsSequence1141 { 3 private static char s[]=new char[105];//接受控制台的输入。 4 private int len;//输入字符串的长度 5 private int dp[][]=new int[105][105];//dp[i][j],从i到j需要插入的最小字符数 6 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 7 private void print(int i,int j){ 8 //输出结果 9 if(i>j) 10 return; 11 if(i==j){ 12 if(s[i]==‘(‘||s[i]==‘)‘) 13 System.out.print("()"); 14 else 15 System.out.print("[]"); 16 } 17 else{ 18 if(pos[i][j]==-1){//如果字符是匹配的 19 System.out.print(s[i]); 20 print(i+1,j-1); 21 System.out.print(s[j]); 22 } 23 else{//字符不匹配 24 print(i,pos[i][j]); 25 print(pos[i][j]+1,j); 26 } 27 } 28 }; 29 private void getDP(){ 30 int inf=0x7fffffff; 31 int i,j,k,mid; 32 int len=s.length; 33 for(i=0;i<len;i++){ 34 for(j=0;j<len;j++){ 35 dp[i][j]=0; 36 } 37 dp[i][i]=1; 38 } 39 for(k=1;k<len;k++){ 40 for(i=0;i+k<len;i++){ 41 j=i+k;//表示不断地分隔,比较不同字符之间是否匹配 42 dp[i][j]=inf;//表示目前不可匹配 43 if((s[i]==‘(‘&&s[j]==‘)‘)||(s[i]==‘[‘&&s[j]==‘]‘)){ 44 //表示字符配对 45 dp[i][j]=dp[i+1][j-1]; 46 pos[i][j]=-1; 47 } 48 //若字符不配对或者其他,表示寻找更少的配对数 49 for(mid=i;mid<j;mid++){ 50 if(dp[i][j]>dp[i][mid]+dp[mid+1][j]){ 51 dp[i][j]=dp[i][mid]+dp[mid+1][j]; 52 pos[i][j]=mid; 53 } 54 } 55 } 56 } 57 } 58 public static void main(String[] args){ 59 Scanner sc = new Scanner(System.in); 60 //String temp=null; 61 BracketsSequence1141 A=new BracketsSequence1141(); 62 s=sc.nextLine().toCharArray(); 63 if(s.length==0){ 64 System.out.println(); 65 } 66 else{ 67 A.getDP(); 68 A.print(0,s.length-1); 69 } 70 } 71 }
1141Brackets Sequence结题报告
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。