http://poj.org/problem?id=2955

 

 

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 … 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 i1i2, …, im where 1 ≤ i1 < i2 < … < im ≤ nai1ai2 … 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

6
6
4
0
6

 

 p[i][j]表示从i到j个可以组成的括号最大值,则若dp[i+1][j]已取到最大值,则dp[i][j] 的取值为 dp[i+1][j] , 或若 s[i] 与 第i+1个到第j个中某个括号匹配(假定为第k个),则有dp[i][j] = max(dp[i+1][j], dp[i+1][k-1] + 2 + dp[k+1][j]) (注:要考虑k == i+1的情况要分开讨论)

 

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

const int INF = 0x3f3f3f3f;
#define N 105

char s[N];
int  dp[N][N];

int main()
{
    while(scanf("%s", s), strcmp(s, "end"))
    {
        int i, j, k, len=strlen(s)-1;

        memset(dp, 0, sizeof(dp));

        for(i=len-1; i>=0; i--)
        {
            for(j=i+1; j<=len; j++)
            {
                dp[i][j] = dp[i+1][j];

                for(k=i+1; k<=j; k++)
                {
                    if((s[i]=='(' && s[k]==')') || (s[i]=='[' && s[k]==']'))
                    {
                        if(k==i+1) dp[i][j] = max(dp[i][j], dp[k+1][j]+2);
                        else       dp[i][j] = max(dp[i][j], dp[i+1][k-1]+dp[k+1][j]+2);
                    }
                }
            }
        }

        printf("%d\n", dp[0][len]);
    }
    return 0;
}

 

记忆化索搜:

(感觉记忆化搜索只是把在递归中已经计算过的值给记录下来, 不知道是否理解有悟,慢慢用吧!!!)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define N 105
#define max(a,b) (a>b?a:b)

char s[N];
int  dp[N][N];

int OK(int L, int R)
{
    if((s[L]=='[' && s[R]==']') || (s[L]=='(' && s[R]==')'))
        return 2;
    return 0;
}

int DFS(int L, int R)
{
    int i;

    if(dp[L][R]!=-1)
        return dp[L][R];
    if(L+1==R)
        return OK(L,R);
    if(L>=R)
        return 0;

    dp[L][R] = DFS(L+1, R);

    for(i=L+1; i<=R; i++)
    {
        if(OK(L,i))
            dp[L][R] = max(dp[L][R], DFS(L+1, i-1)+DFS(i+1, R)+2);
    }
    return dp[L][R];
}

int main()
{
    while(scanf("%s", s), strcmp(s, "end"))
    {
        memset(dp, -1, sizeof(dp));
        printf("%d\n", DFS(0, strlen(s)-1));
    }
    return 0;
}

 

posted on 2015-12-16 16:52  栀蓝  阅读(171)  评论(0编辑  收藏  举报

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