动态规划——J 括号配对问题

J - 括号匹配

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

题目大意：就是寻找有多少对括号匹配，并输出它们的个数

解题思路：

两种括号的匹配问题，这个问题不能用顺序解决，因为两种括号存在交叉的情形，之前我们用队列的方法做过，而且有些情况并不能知道优先匹配哪种括号，但是今天不用队列方法，用DP

这个题目注意分段max(d[j][i+j-1],d[j][k]+d[k+1][j+i-1])

程序代码：

#include <cstdio>
#include <cstring>
using namespace std;
char s[110];
int d[110][110];
bool match(char a, char b)
{
  return (a == '(' && b == ')') || (a == '[' && b == ']');
}
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    while(~scanf("%s",s))
    {
        if(strcmp(s,"end")==0) break;
        int len=strlen(s);
        for(int i=0;i<len;i++)
        {
            d[i][i]=0;
            if(match(s[i],s[i+1]))
                d[i][i+1]=2;
            else
                d[i][i+1]=0;
        }
        for(int i=3;i<=len;i++)
            for(int j=0;j+i-1<len;j++)
        {
            d[j][i+j-1]=0;
            if(match(s[j],s[i+j-1]))
                d[j][i+j-1]=d[j+1][i+j-2]+2;
            for(int k=j;k<i+j-1;k++)
               d[j][i+j-1]=max(d[j][i+j-1],d[j][k]+d[k+1][j+i-1]);
        }
            printf("%d
",d[0][len-1]);
    }
      return 0;
}