poj 1114

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

()[()]

Source

Northeastern Europe 2001

状态：dp[i][j]表示 i 到 j 最多的匹配个数
转移方程：dp[i][j] = max（dp[i][j], dp[i][k] + dp[k+1][j]);

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N = 105;
char str[N];
int dp[N][N];
bool judge(char a, char b)
{
    if(a == ‘(‘ && b == ‘)‘) return true;
    if(a == ‘[‘ && b == ‘]‘) return true;
    return false;
}
int main()
{
    while(gets(str) != NULL)
    {
        if(!strcmp(str, "end")) break;
        int len = strlen(str);
        for(int i = 0; i < len; i++)
        {
            dp[i][i] = 0;
            if(judge(str[i], str[i+1]))
                dp[i][i+1] = 2;
            else
                dp[i][i+1] = 0;
        }
        for(int k = 2; k < len; k++) //枚举子串的长度
        {
            for(int i = 0; i + k < len; i++)
            {
                int r = i + k;
                dp[i][r] = 0;
                if(judge(str[i], str[r]))
                    dp[i][r] = dp[i+1][r-1] + 2;
                for(int j = i; j < r; j++)
                    dp[i][r] = max(dp[i][r], dp[i][j] + dp[j+1][r]);
            }
        }
        printf("%d\n",dp[0][len-1]);
    }
    return 0;
}