"POJ 1141" Brackets Sequence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27996		Accepted: 7936		Special Judge

Description

Let us define a regular brackets sequence in the following by:

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 is regular brackets sequences:

(), [], (()), ([]), ()[], ()[()]

And all of the following character sequences is not:

(, [, ), ) (, ([)], ([(]

Some sequence of characters ' (', ') ', ' [', and '] ' is given. You is to find the shortest possible regular brackets sequence, that contains the given character sequence as a Subsequen Ce. Here, a string A1 A2 ... is called a subsequence of the string B1 b2 ... bm, if there exist such indices 1 = I1 < I2 < ... < in = m, which AJ = bij for all 1 = j = N.

Input

The input file contains at most of brackets (characters ' (', ') ', ' [' and '] ') that is situated on a single line without Any other characters among them.

Output

Write to the output file A contains some regular brackets sequence this has the minimal possible length a nd contains the given sequence as a subsequence.

Sample Input

([(]

Sample Output

()[()]

Source

Northeastern Europe 2001 can no longer believe POJ special judge, in fact there is no good felling. Got me on my knees ... A test data to see ... Cough cough. Well, the answer to this question, in fact, is the interval dp+ answers output. The recursive output of the answer is also relatively straightforward. The DP equation can be found in the Rujia Teacher's Black book, which is not mentioned here.

#include <cstdio>#include<cstring>using namespacestd;Chars[ the];intN, f[ the][ the], p[ the][ the];voidPrintintLintR) {    if(L > R)return; if(P[l][r] = =-1) {printf ("%c", S[l]); Print (L+1, R-1); printf ("%c", S[r]); return; }    if(P[l][r] = =-2) {printf ("%c", S[l]); Print (L+1, R); if(S[l] = ='(') printf (")"); Elseprintf"]"); return; }    if(P[l][r] = =-3)    {        if(S[r] = =')') printf ("("); Elseprintf"["); Print (L, R-1); printf ("%c", S[r]); return; } print (L, P[l][r]-1); Print (P[l][r], R);}intMain () {fgets (s),104, stdin); N= strlen (s)-1; Memset (F,0x7f,sizeof(f)); Memset (P,0,sizeof(p));  for(inti =0; I < n; ++i) {F[i][i]=1; if(S[i] = ='('|| S[i] = ='[') P[i][i] =-2; ElseP[i][i] =-3;  for(intj =0; J < I; ++j) F[i][j]=0; }     for(inti = n-1; I >=0; --i) for(intj = i +1; J < N; ++j) {if((s[i] = ='('|| S[J] = ='[') && F[i][j] > f[i +1][J] +1) {F[i][j]= F[i +1][J] +1; P[I][J]= -2; }            if((s[j] = =')'|| S[J] = =']') && F[i][j] > F[i][j-1] +1) {F[i][j]= F[i][j-1] +1; P[I][J]= -3; }            if((s[i] = ='('&& S[j] = =')') || (S[i] = ='['&& S[j] = =']')) && F[i][j] > f[i +1][j-1]) {F[i][j]= F[i +1][j-1]; P[I][J]= -1; }             for(intK = i +1; K <= J; ++k)if(F[i][j] >= f[i][k-1] +F[k][j]) {F[i][j]= F[i][k-1] +F[k][j]; P[I][J]=K; }} print (0N1); printf ("\ n"); return 0;}

"POJ 1141" Brackets Sequence