POJ 2955 Brackets （区间DP）

题目链接：http://poj.org/problem?id=2955

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1977		Accepted: 1012

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

 

合法序列就是括号可以两两匹配的。

思路就是区间DP的思想了。

我的代码是采用记忆化搜索写出来的。

 

状态转移方程dp[i][j]=max(dp[i+1][j],2+dp[i+1][k-1]+dp[k+1][j])       i和j是一对括号 && i<k<=j

其实就是看第i个括号的情况。

舍弃第i个括号，就是dp[i+1][j],或者是往后找和i匹配的，然后就分成了两部分了。

//============================================================================

// Name        : POJ.cpp

// Author      : 

// Version     :

// Copyright   : Your copyright notice

// Description : Hello World in C++, Ansi-style

//============================================================================



#include <iostream>

#include <string.h>

#include <stdio.h>

#include <algorithm>

using namespace std;

const int MAXN=110;

char str[MAXN];

int dp[MAXN][MAXN];

int solve(int i,int j)

{

    if(dp[i][j]!=-1)return dp[i][j];

    if(j<=i)return dp[i][j]=0;

    else if(j==i+1)

    {

        if( (str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']') )return dp[i][j]=2;

        else return dp[i][j]=0;

    }

    dp[i][j]=solve(i+1,j);

    for(int k=i+1;k<=j;k++)

        if( (str[i]=='('&&str[k]==')')||(str[i]=='['&&str[k]==']') )

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

    return dp[i][j];

}



int main()

{

//    freopen("in.txt","r",stdin);

//    freopen("out.txt","w",stdout);

    while(scanf("%s",str)==1)

    {

        if(strcmp(str,"end")==0)break;

        memset(dp,-1,sizeof(dp));

        int n=strlen(str);

        printf("%d\n",solve(0,n-1));

    }

    return 0;

}