POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25132		Accepted: 7083		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

()[()]

#include <iostream>
#include <stack>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;

#define INF 0x3fffffff
#define N 105
#define ms(a, val) memset(a, val, sizeof(a))
int dp[N][N], p[N][N];
string s;
/*
if(s[i,j] == "[]" || "()")
    dp[i][j] = dp[i+1][j-1]
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j]) i<=k<j
*/
void print(int i, int j)
{
    if (i > j)
    {
        return;
    }
    if (i == j)
    {
        if (s[i] == '(' || s[j] == ')')
            cout << "()";
        else
            cout << "[]";
        return;
    }
    if (p[i][j] == -1)
    {
        cout << s[i];
        print(i + 1, j - 1);
        cout << s[j];
    }
    else
    {
        print(i, p[i][j]);
        print(p[i][j] + 1, j);
    }
}
int main()
{
    int n, t;
    while (getline(cin, s))
    {
        n = s.length();
        ms(dp, 0);
        for (int i = 0; i < n; i++)
        {
            dp[i][i] = 1;
        }
        for (int m = 1; m < n; m++)
        {
            for (int i = 0, j = m; j < n; i++, j++)
            {
                dp[i][j] = INF;
                p[i][j] = -1;
                if ((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']'))
                {
                    dp[i][j] = dp[i + 1][j - 1];
                }
                for (int k = i; k < j; k++)
                {
                    t = dp[i][k] + dp[k + 1][j];
                    if (t < dp[i][j])
                    {
                        dp[i][j] = t;
                        p[i][j] = k;
                    }
                }
            }
        }
        print(0, n - 1);
        cout << endl;
    }
    return 0;
}