POJ 2955-Brackets（括號匹配-區間DP）

kewlgrl發表於2016-05-28
Racket

Brackets

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 5484   Accepted: 2946

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 i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … 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

Source

Stanford Local 2004


題目意思：

最大括號匹配數。

解題思路：

From

定義dp [ i ] [ j ] 為串中第 i 個到第 j 個括號的最大匹配數目

那麼我們假如知道了 i 到 j 區間的最大匹配，那麼i+1到 j+1區間的是不是就可以很簡單的得到。

那麼 假如第 i 個和第 j 個是一對匹配的括號那麼 dp [ i ] [ j ] = dp [ i+1 ] [ j-1 ] + 2 ;

那麼我們只需要從小到大列舉所有 i 和 j 中間的括號數目，然後滿足匹配就用上面式子dp，然後每次更新dp [ i ] [ j ]為最大值即可。

更新最大值的方法是列舉 i 和 j 的中間值，然後讓  dp[ i ] [ j ] = max ( dp [ i ] [ j ] , dp [ i ] [ f ] + dp [ f+1 ] [ j ] ) 。 



#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 202
#define INF 999999
int dp[MAXN][MAXN];

int main()
{
    string s;
    while(cin>>s)
    {
        if(s=="end")  break;
        memset(dp,0,sizeof(dp));
        int i,j,k,f,len=s.size();
        for(i=1; i<len; i++)
            for(j=0,k=i; k<len; j++,k++)
            {
                if((s[j]=='('&&s[k]==')')||(s[j]=='['&&s[k]==']'))
                    dp[j][k]=dp[j+1][k-1]+2;
                for(f=j; f<k; f++)
                    dp[j][k]=max(dp[j][k],dp[j][f]+dp[f+1][k]);
            }
        //len-dp[0][len-1]表示需要補充多少括號才能都匹配成功
        cout<<dp[0][len-1]<<endl;
    }
    return 0;
}


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