POJ1141 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

()[()]

Source

Northeastern Europe 2001

正解：DP

解题报告：

　　DP题，乍一看我居然不会做，也是醉了。开始想用贪心水过，发现会gi烂。

　　详细博客：http://blog.csdn.net/lijiecsu/article/details/7589877

　　不详细说了，见代码：

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<vector>
 using namespace std;
 const int MAXN = ;
 char ch[MAXN];
 int l;
 int f[MAXN][MAXN],c[MAXN][MAXN];

 inline void output(int l,int r){
     if(l>r) return ;
     if(l==r) {
     if(ch[l]=='(' || ch[l]==')') printf("()");
     else printf("[]");
     }
     else{
     if(c[l][r]>=) {
         output(l,c[l][r]);
         output(c[l][r]+,r);
     }
     else{
         if(ch[l]=='(') {
         printf("(");
         output(l+,r-);
         printf(")");
         }
         else{
         printf("[");
         output(l+,r-);
         printf("]");
         }
     }
     }
 }

 inline void solve(){
     scanf("%s",ch);
     int len=strlen(ch);
     for(int i=;i<len;i++) f[i][i]=;
     for(int i=;i<len;i++) for(int j=;j<len;j++) c[i][j]=-;
     for(int l=;l<=len-;l++)
     for(int i=;i+l<=len-;i++){
         int j=i+l;
         int minl=f[i][i]+f[i+][j];
         c[i][j]=i;
         for(int k=i+;k<j;k++){
         if(minl>f[i][k]+f[k+][j]) {
             minl=f[i][k]+f[k+][j];
             c[i][j]=k;
         }
         }
         f[i][j]=minl;

         if(( ch[i]=='(' && ch[j]==')' ) || ( ch[i]=='[' && ch[j]==']' )) {
         if(f[i][j]>f[i+][j-]) {
             f[i][j]=f[i+][j-];
             c[i][j]=-;
         }
         }
     }

     output(,len-);
     printf("\n");
 }

 int main()
 {
     solve();
     return ;
 }