POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13305		Accepted: 3585		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

Northeastern Europe 2001

/*

http://acm.pku.edu.cn/JudgeOnline/problem?id=1141

动态规划，与最优二叉查找树类似，从下往上，从小到大，遍历长度len，然后遍历位置k

*/

#include <iostream>

#define MAX_N 100
using namespace std;

char input[MAX_N + 1];
int len;
int pos[MAX_N + 1][MAX_N + 1];
int minV[MAX_N + 1][MAX_N + 1];

void print(int f, int t)
{
    if(f > t)
        return;
    if(pos[f][t] == -1)
    {
        cout<<input[f];
        print(f + 1, t - 1);
        cout<<input[t];
    }
    else if(f == t)
    {
        char type = input[f];
        if(type == ')' || type == '(')
            cout<<"()";
        else
            cout<<"[]";
    }
    else
    {
        print(f, pos[f][t]);
        print(pos[f][t] + 1, t);
    }
}

bool match(char t1, char t2)
{
    return ((t1 == '(' && t2 == ')') || (t1 == '[' && t2 == ']'));
}
int main()
{
    char ch;
    len = 0;
    while((ch = getchar()) != 10)
    {
        input[++len] = ch;
    }
    int i, j;
    for(i = 1; i <= len; i++)
    {
        for(j = 1; j <= len; j++)
        {
            if(i == j)
                minV[i][j] = 1;
        }
    }
    int l, k;
    for(l = 2; l <= len; l++)
    {
        for(i = 1; i <= len - l + 1; i++)
        {
            j = i + l - 1;
            int minVal = INT_MAX, minPos = 0, curVal;
            for(k = i; k < j; k++)
            {
                curVal = minV[i][k] + minV[k + 1][j];
                if(curVal < minVal)
                {
                    minVal = curVal;
                    minPos = k;
                }
            }
            if(match(input[i], input[j]))
            {
                curVal = minV[i + 1][j - 1];
                if(curVal < minVal)
                {
                    minVal = curVal;
                    minPos = -1; //min trim head and tail
                }
            }
            minV[i][j] = minVal;
            pos[i][j] = minPos;
           
        }
    }
    print(1, len);
    cout<<endl;
    return 0;
}