Zoj1463:brackets Sequence (Gap DP)

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.

This problem contains multiple test cases!

The first line of a multiple input was an integer N and then a blank line followed by N input blocks. Each input block was in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1

([(]

Sample Output

()[()]

The key is in the input and output formats. God pit.

Interval Dp,dp[i][j] Indicates the number of matches between the interval I to J, whether the characters at both ends of the interval match exactly, and if one dp[i][j] = max (dp[i][k]+dp[k+1][j],dp[i+1][j-1]+1) is able to match the state transfer, If not match is dp[i][j] = max (dp[i][j],dp[i][k]+dp[k+1][j]);

#include <stdio.h> #include <string.h> #include <algorithm>using namespace std; #define UP (I,x,y) for ( i=x;i<=y;i++) #define DOWN (i,x,y) for (i=x;i>=y;i--) #define MEM (A, B) memset (A,b,sizeof (a)) #define W (a) while (a) Char str[105];int t,len,dp[105][105],mark[105][105],pos[105];void dfs (int i,int j) {if (mark[i][j]==-1) {pos[        I]=pos[j]=1;    DFS (I+1,J-1);        } else if (mark[i][j]>=0) {DFS (i,mark[i][j]);    DFS (MARK[I][J]+1,J); } return;    int main () {int l,i,j,k;    scanf ("%d%*c%*c", &t);        while (t--) {gets (str);        Len=strlen (str);            if (!len) {printf ("\ n");            if (t) printf ("\ n");        Continue            } up (i,0,len-1) up (j,0,len-1) {mark[i][j]=-2;        dp[i][j]=0;        } mem (pos,0);        i=j=l=0;                W (L<len) {if (i==j) {i++,j++;         if (J==len)           I=0,l++,j=l;            Continue } if ((str[i]== ' (' &&str[j]== ') ') | |            (str[i]== ' [' &&str[j]== '])                        {Up (k,i,j-1) {if (Dp[i][j]<dp[i][k]+dp[k+1][j]) {                        Mark[i][j]=k;                    DP[I][J]=DP[I][K]+DP[K+1][J];                    }} if (dp[i][j]<dp[i+1][j-1]+1) {mark[i][j]=-1;                dp[i][j]=dp[i+1][j-1]+1; }} else {up (k,i,j-1) {if (dp[i][j]<                        Dp[i][k]+dp[k+1][j]) {mark[i][j]=k;                    DP[I][J]=DP[I][K]+DP[K+1][J];            }}} i++,j++;                if (J==len) {l++;                i=0;            J=l;   }     } dfs (0,len-1);            Up (i,0,len-1) {if (pos[i]==1) printf ("%c", Str[i]); else if (str[i]== ' (' | | |            str[i]== ') printf ("()");        else printf ("[]");        } printf ("\ n");            if (t) {printf ("\ n");        GetChar (); }} return 0;}

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Zoj1463:brackets Sequence (Gap DP)