poj2955 Brackets

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

66406

题意：最大括号匹配的一个长度。

区间dp，卡它的那个要求。

dp[i][j]表示i-j的最大值。

然后如果i和j匹配的话，那么就是dp[i+1][j-1]+2。

然后就是第二种情况，枚举中间值k,则有dp[i][k] + dp[k+1][j]。

两种情况取一个大的就可以了。

#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>using namespace std;typedef long long LL;const int inf = 1e9;const int MAXN = 100+7;int n,m;char s[MAXN];int dp[MAXN][MAXN];int main(){   while(1)   {       memset(s,0,sizeof s);       memset(dp,0,sizeof dp);       scanf("%s",s+1);       if(s[1] == 'e')break;       int l = strlen(s+1);       for(int i = l ; i >= 1 ; --i)        for(int j = i+1 ; j <= l ; ++j)       {            if(s[j] - s[i] == 1 || s[j] - s[i] == 2)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[1][l]);   }}