[DP] UVA-1626 Brackets sequence

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 a₁a₂ . . . a_n is called a subsequence of the string b₁b₂ . . . b_m, if there exist such indices 1 ≤ i₁ < i₂ < . . . < i_n ≤ m, that a_j = b_ij for all 1 ≤ j ≤ n.

 

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The input file contains at most 100 brackets (characters ‘(’, ‘)’, ‘[’ and ‘]’) that are situated on a single line without any other characters among them.

 

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

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

1

([(]

 

Sample Output

()[()]

题意：

定义如下序列为合法的括号序列：

1、空序列为合法序列

2、若S是合法序列，则(S), [S]也为合法序列

3、若A、B均为合法序列，则AB也为合法序列

题目给出一个括号序列，添加最少的(, ), [, ]使序列合法。

思路：本题是经典的区间DP，思想是不断分割小区间，直到出现(S)或[S]时，状态转移到S，即从dp[i][j]转移到dp[i + 1][j - 1]。

PS：该题的读入和输出很坑，需要注意

#include<cstdio>
#include<cstring>
using namespace std;
int opt[501][501],f[501][501];
char str[501];
void print(int left,int right)
{
    if (left > right) return;
    else
    if (left == right)
    {
        if ((str[left] == '[') || (str[right] == ']')) printf("[]");
        else printf("()");    
    }
    else
        if (f[left][right] == -1)
        {
            printf("%c",str[left]);
            print(left + 1,right - 1);
            printf("%c",str[right]);
        }
        else
        {
            print(left,f[left][right]);
            print(f[left][right] + 1,right);
        }
}
int main()
{
    int t,len,i,j,k,l,Case;
    char s[101],ch;
    scanf("%d",&t);
    scanf("%c",&ch);
    for (Case = 1;Case <= t;Case++)
    {
        scanf("%c",&ch);
        gets(str);
        len = strlen(str);
        if (len == 0)
        {
            printf("
");
            if (Case != t) printf("
");
            continue;
        }
        memset(opt,0,sizeof(opt));
        memset(f,0,sizeof(f));
        for (i = 0;i < len;i++) opt[i][i] = 1;
        for (l = 1;l < len;l++)
        {
            for (i = 0;i < len - l;i++)
            {
                j = i + l;
                opt[i][j] = 999999999;
                if (((str[i] == '(') && (str[j] == ')')) || ((str[i] == '[') && (str[j] == ']')))
                {
                    opt[i][j] = opt[i + 1][j - 1];
                    f[i][j] = -1;
                }
                for (k = i;k < j;k++)
                {
                    if (opt[i][j] > opt[i][k] + opt[k + 1][j])
                    {
                        opt[i][j] = opt[i][k] + opt[k + 1][j];
                        f[i][j] = k;
                    }
                }
            }
        }
        print(0,len - 1);
        printf("
");
        if (Case != t) printf("
");
    }
}