2015 HUAS Summer Trainning #5~J

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
解题思路：题目的意思是在输入的字符串中寻找有多少对括号匹配，并输出它们的个数。需要注意的是，这个问题不能用顺序解决，因为这是两种括号的匹配问题，会出现交叉情况，之前也用过的队列解决过。

程序代码：

#include <stdio.h> 
#include <string.h> 
#include <algorithm> 
using namespace std;

int max(int x,int y)
{
 if(x>y) return x;
 else return y;
}

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; 
}