POJ2955：Brackets(区间DP)

Description

We give the following inductive definition of a “regular brackets” sequence:

• the empty sequence is a regular brackets sequence,
• if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
• if a and b are regular brackets sequences, then ab is a regular brackets sequence.
• no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while the following character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a₁a₂ … a_n, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i₁, i₂, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

Input

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

Output

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

Sample Input

((()))

()()()

([]])

)[)(

([][][)

end

Sample Output

6

6

4

0

6

题意：求出互相匹配的括号的总数

思路：一道区间DP，dp[i][j]存的是i~j区间内匹配的个数

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;



int check(char a,char b)

{

    if(a=='(' && b==')')

        return 1;

    if(a=='[' && b==']')

        return 1;

    return 0;

}



int main()

{

    char str[105];

    int dp[105][105],i,j,k,len;

    while(~scanf("%s",str))

    {

        if(!strcmp(str,"end"))

            break;

        len = strlen(str);

        for(i = 0; i<len; i++)

        {

            dp[i][i] = 0;

            if(check(str[i],str[i+1]))

                dp[i][i+1] = 2;

            else

                dp[i][i+1] = 0;

        }

        for(k = 3; k<=len; k++)

        {

            for(i = 0; i+k-1<len; i++)

            {

                dp[i][i+k-1] = 0;

                if(check(str[i],str[i+k-1]))

                    dp[i][i+k-1] = dp[i+1][i+k-2]+2;

                for(j = i; j<i+k-1; j++)

                    dp[i][i+k-1] = max(dp[i][i+k-1],dp[i][j]+dp[j+1][i+k-1]);

            }

        }

        printf("%d\n",dp[0][len-1]);

    }



    return 0;

}