POJ 2955 Brackets (动规)

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

Source

Stanford Local 2004

?如果找到一对匹配的括号[xxx]oooo，就把区间分成两部分，一部分是xxx，一部分是oooo，然后以此递归直到区间长度为1或者为2

?dp[i][j]表示区间[i,j]之间的最长合法括号子序列长度

?dp[i][j] = min{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)

}

代码：

#include <iostream>
#include <string.h>
#include <math.h>
#include <stdio.h>
using namespace std;
#define M 110
int dp[M][M];
char str[M];
int math(int x,int y)      //判断是否配对。
{
    return str[x]==‘(‘ && str[y]==‘)‘||
           str[x]==‘{‘ && str[y]==‘}‘||
           str[x]==‘[‘ && str[y]==‘]‘;
}
int DFS(int s,int e)
{
    int i,ret,t;
    if(e<s)                return 0;                //对各种特殊情况进行考虑。
    if(e==s)               return dp[s][e]=0;
    if(e-s==1)             return dp[s][e]=math(s,e);
    if(dp[s][e]!=-1)       return dp[s][e];        //如果这个状态已经计算过。
    ret=DFS(s+1,e);                                //姿势不对，还是看别人的，不懂。
    for(i=s+1;i<=e;i++)                            //找能配对的括号。
        if(math(s,i))
    {
        t=DFS(s+1,i-1)+DFS(i+1,e)+1;
        if(t>ret) ret=t;
    }
    return dp[s][e]=ret;
}
int main()
{
    int i,j,k,n;
    while(scanf("%s",str))
    {
        if(str[0]==‘e‘) break;
        n=strlen(str);
        memset(dp,-1,sizeof(dp));
        DFS(0,n-1);
        printf("%d\n",dp[0][n-1]*2);
    }
    return 0;
}

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