Poj 1141 brackets Sequence

Brackets Sequence

Time limit:1000 ms		Memory limit:65536 K
				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

()[()]

Source

The path of the DP record in the northeastern Europe 2001 range, which is combined with the matching of the brackets in the previous two questions + modifying the echo string .. Note empty rows

 1 #include<set> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 const int N = 110; 9 #define For(i,n) for(int i=1;i<=n;i++)10 #define Rep(i,l,r) for(int i=l;i<=r;i++)11 #define Down(i,r,l) for(int i=r;i>=l;i--)12 13 struct State{14     int v,op;15 }dp[N][N];16 17 char st[N];18 int n;19 20 bool match(int x,int y){21     if(st[x]==‘(‘) return (st[y]==‘)‘);22     if(st[x]==‘[‘) return (st[y]==‘]‘);23     return false;24 }25 void DP(){26     n=strlen(st+1);27     For(i,n)28       For(j,n)29         if(j<i)  dp[i][j].v=0;30         else     dp[i][j].v=214748364;31     For(i,n) dp[i][i].v=1;32     Down(i,n-1,1)33       Rep(j,i,n){34           if(match(i,j))35             if(dp[i][j].v>dp[i+1][j-1].v){36                 dp[i][j].v=dp[i+1][j-1].v;37                 dp[i][j].op=0;38             }39           Rep(k,i,j-1)40             if(dp[i][k].v+dp[k+1][j].v<dp[i][j].v){41                 dp[i][j].v=dp[i][k].v+dp[k+1][j].v;42                 dp[i][j].op=k;43             }44       }45 }46 47 void Print(int l,int r){48     if(l>r) return;49     if(l==r){50         if(st[l]==‘(‘||st[l]==‘)‘) printf("()");51         if(st[l]==‘]‘||st[l]==‘[‘) printf("[]");52         return;53     }54     if(dp[l][r].op){55         Print(l,dp[l][r].op);56         Print(dp[l][r].op+1,r);57     }58     else{59         printf("%c",st[l]);60         Print(l+1,r-1);61         printf("%c",st[r]);62     }63 }64 65 int main(){66     while(gets(st+1)){67         if(strlen(st+1)){68             DP();69             Print(1,n);70         }71         puts("");72     }73     return 0;74 }

 

Poj 1141 brackets Sequence