POJ 2955 Brackets （区间DP）

最新推荐文章于 2020-06-03 17:42:50 发布

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最新推荐文章于 2020-06-03 17:42:50 发布

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本文探讨了一种使用区间动态规划解决寻找字符串中最长有效括号子序列的问题，通过优化的递推关系处理特殊情况，实现了高效求解。

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题目链接：POJ 2955

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 … an, 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, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim 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

题意：一个串中含有‘ ( ’, ‘ ) ’, ‘ [ ’, ‘ ] ’ 四种字符，两种括号对"()", "[]"可以计数，括号对两边同时加上同一种左右括号也可以计数，即：(()), [()]。要求输出一个串中可以计数的字符长度。

题解：这个题用区间DP来求解，仍用DP[i][j]来存i 到j 的最大匹配值，寻找最优切割点dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j])也是中规中矩的，只不过某种情况下DP[i][j]会因为i 和 j 的位置存在括号匹配的情况增加，比如"(())", 但有时两边的括号已经跟别的括号匹配了，比如“()()", 这种情况下就需要一个另外的判断式子dp[i][j] = max(dp[i][j], dp[i + 1][j - 1] + 1），用来处理这种情况。

AC代码：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int i, j, n, ans, len, k;
    int dp[105][105] = {0};
    char str[105];
    while(scanf("%s", str) && str[0] != 'e'){
        for(len = 2;len <= strlen(str);len++){//区间长度
            for(i = 0;i < strlen(str);i++){//枚举起点i
                j = i + len - 1;//计算终点j
                if(j >= strlen(str))//越界跳出
                    break;
                for(k = i;k <= j;k++)
                    dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j]);//递推关系， 通过小区间计算得出大区间的值
                if(str[i] == '(' && str[j] == ')' || str[i] == '[' && str[j] == ']')
                    dp[i][j] = max(dp[i][j], dp[i + 1][j - 1] + 1);//处理特殊情况
            }
        }
        printf("%d\n", dp[0][strlen(str) - 1] * 2);
        memset(dp, 0, sizeof(dp));
        memset(str, 0, sizeof(str));
    }
    return 0;
}