POJ 1141 - Brackets Sequence

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

()[()]

题意：给出一个括号序列，求出添加括号数最少的情况下，能得到的括号序列。

记录下断点和最小括号数，并且输出时用递归方法。

#include <cstdio>#include <cstring>int ans[105][105];int itv[105][105];char str[1005];int len;void print(int s, int e){    if (s > e)        return;    if (s == e)    {        if (str[s] == '(' || str[s] == ')')            printf("()");        else            printf("[]");    }    else    {        if (itv[s][e] >= 0)        {            print(s, itv[s][e]);            print(itv[s][e]+1, e);        }        else        {            if (str[s] == '(')            {                printf("(");                print(s+1, e-1);                printf(")");            }            else            {                printf("[");                print(s+1, e-1);                printf("]");            }        }    }}int main(){    gets(str);        len = strlen(str);        for (int i = 0; i < len; ++i)        {            for (int j = 0; j < len; ++j)            {                ans[i][i] = 1;                itv[i][j] = -1;            }        }        int minn;        for (int t = 1; t < len; ++t)        {            for (int i = 0; i + t < len; ++i)            {                int j = i + t;                ans[i][j] = 0x3fff;                minn = ans[i][i] + ans[i+1][j];                itv[i][j] = i;                for (int k = i+1; k < j; ++k)                {                    if (ans[i][k] + ans[k+1][j] < minn)                    {                        minn = ans[i][k] + ans[k+1][j];                        itv[i][j] = k;                    }                }                ans[i][j] = minn;                if ((str[i] == '(' && str[j] == ')') || (str[i] == '[' && str[j] == ']'))                {                    if (ans[i+1][j-1] < minn)                    {                        ans[i][j] = ans[i+1][j-1];                        itv[i][j] = -1;                    }                }            }        }        print(0, len - 1);        printf("\n");    return 0;}