POJ 1141 Brackets Sequence 动态规划

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

Northeastern Europe 2001

    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;
}