【POJ

L - 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：给出一个由'[' , ']' , '(' , ')' 构成的字符串，统计括号匹配的个数。

分析：一道基础的区间dp，先将dp数组初始化为0。然后遍历起点i，终点j，中间点k，要注意处理边界s[i]和s[j]是否匹配。

状态转移方程：dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);

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

代码如下：

#include <set>#include <map>#include <queue>#include <cmath>#include <vector>#include <cctype>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define mod 835672545#define INF 0x3f3f3f3f#define LL long longusing namespace std;const int MX = 105;char s[MX];int dp[MX][MX];int main(){    while(~scanf("%s", s)){        if(s[0] == 'e') return 0;        memset(dp, 0, sizeof(dp));        int len = strlen(s);        for(int i = len-2; i >= 0; i--){            for(int j = i+1; j <= len-1; j++){                for(int k = i; k < j; k++){                    dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);                    if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')){                        dp[i][j] = max(dp[i][j], dp[i+1][j-1] + 2);                    }                }            }        }        printf("%d\n", dp[0][len-1]);    }    return 0;}