POJ 2955 Brackets

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6622		Accepted: 3558

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

Source

Stanford Local 2004

 

 

题目大意：给你一个长度不超过100的括号序列，求最长合法括号子序列的长度。合法的括号序列满足下列条件：

1.空的括号序列是合法的；

2.如果一个括号序列s是合法的，那么(s)和[s]都是合法的；

3.如果两个括号序列a和b都是合法的，那么ab也是合法的；

4.其他的括号序列都是不合法的。

例如：(), [], (()), ()[], ()[()]都是合法的，而(, ], )(, ([)], ([(]都是不合法的。

 

解题思路：一道典型的区间DP模型题目。分析一下问题，可以发现：如果找到一对匹配的括号，例如[xxx]oooo，就把区间分成两部分，一部分是xxx，另一部分是oooo。

设dp[i][j]表示区间[i,j]之间的最长合法括号子序列的长度，那么当i<j时，如果区间[i+1,j]内没有与i匹配的括号，则dp[i][j]=dp[i+1][j]；如果存在一个与之匹配的k，那么dp[i][j]=max{dp[i+1][j], dp[i+1][k-1]+dp[k+1][j]+1(i<=k<=j&&i和k是一对匹配的括号)}。因此，我们将整个串长作为区间进行搜索，那么最后2*dp[0][len-1]即为答案，len表示串的长度。详见代码。

 

 

附上AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 105;
char str[maxn];
int dp[maxn][maxn];

bool match(char a, char b){
    return (a=='('&&b==')') || (a=='['&&b==']');
}

int dfs(int l, int r){
    if (l > r)
        return 0;
    if (l == r)
        return dp[l][r] = 0;
    if (l+1 == r)
        return dp[l][r] = match(str[l], str[r]);
    if (dp[l][r] != -1)
        return dp[l][r];
    int ans = dfs(l+1, r);
    for (int i=l; i<=r; ++i)
        if (match(str[l], str[i]))
            ans = max(ans, dfs(l+1, i-1)+dfs(i+1, r)+1);
    return dp[l][r] = ans;
}

int main(){
    while (~scanf("%s", str) && str[0]!='e'){
        memset(dp, -1, sizeof(dp));
        int len = strlen(str);
        dfs(0, len-1);
        printf("%d
", 2*dp[0][len-1]);
    }
    return 0;
}