## POJ 1141 Brackets Sequence 动态规划

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

`()[()]`

Source

设dp[i,j]为从位置i到位置j需要加入字符的最小次数，有dp[i,j]=min(dp[i,k]+dp[k+1,j])，其中i<=k<j。特别的当s[i]='[' s[j]=']'或者s[i]='(' s[j]=')'时，dp[i,j]=dp[i+1,j-1]。初始条件为dp[i,i]=1，其中0<=i<len。 #include <iostream> using namespace std;  const int MAXN = 110; char str[MAXN]; int dp[MAXN][MAXN],path[MAXN][MAXN];   void output(int i,int j){ if(i>j) return;  if(i==j){ if(str[i]=='[' || str[i]==']') printf("[]"); else printf("()"); }  else if(path[i][j]==-1){ printf(
"%c",str[i]); output(i
+1,j-1); printf(
"%c",str[j]); }  else{ output(i,path[i][j]); output(path[i][j]
+1,j); } }  int main(){ int i,j,k,r,n;  while(gets(str)){ n
=strlen(str);  if(n==0){ printf(
"\n"); continue; } memset(dp,
0,sizeof(dp)); for(i=0;i<n;i++) dp[i][i]=1; for(r=1;r<n;r++)  for(i=0;i<n-r;i++){ j
=i+r; dp[i][j]
=INT_MAX; if((str[i]=='(' && str[j]==')')||(str[i]=='[' && str[j]==']')) if(dp[i][j]>dp[i+1][j-1]) dp[i][j]
=dp[i+1][j-1],path[i][j]=-1; for(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; } output(
0,n-1); printf(
"\n"); } return 0; }

posted on 2009-06-29 11:31 极限定律 阅读(2508) 评论(0)  编辑 收藏 引用 所属分类: ACM/ICPC < 2015年7月 >
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