ACM: 简单动态规划题 poj 2955

                                            Brackets

 

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: 计算最大的括号匹配数目.

 

解题思路:

       1. dp[i][j]: 第i个字符到到第j个字符最大的匹配数.

       2. (str[i] == '(' && str[j] == ')') || (str[i] == '[' && str[j] == ']')

            dp[i][j] = max(dp[i][j] , dp[i+1][j-1]+2);

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

 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 105

char str[MAX];
int dp[MAX][MAX];
int len;

inline int max(int a,int b)
{
 return a > b ? a : b;
}

int main()
{
// freopen("input.txt","r",stdin);
 while(scanf("%s",str) != EOF)
 {
  if(strcmp(str,"end") == 0)
   break;

  memset(dp,0,sizeof(dp));
  len = strlen(str);
  for(int p = 0; p < len; ++p)
  {
   for(int i = 0, j = p; j < len; ++i, ++j)
   {
    if( (str[i] == '(' && str[j] == ')') || (str[i] == '[' && str[j] == ']') )
     dp[i][j] = max(dp[i][j],dp[i+1][j-1]+2);
    for(int k = i; k < j; ++k)
     dp[i][j] = max(dp[i][j],dp[i][k]+dp[k+1][j]);
   }
  }
  printf("%d\n",dp[0][len-1]);
 }
 return 0;
}