Poj 2955 brackets（区间dp）

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7795		Accepted: 4136

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

Source

Stanford Local 2004

题解：

f[i][j]表示i到j的最大括号匹配数

#include <iostream>

#include<cstdio>

#include<cstring>

#include<map>

#include<set>

#include<queue>

#include<vector>

#include<deque>

#include<algorithm>

#include<string>

#include<stack>

#include<cmath>

using namespace std;

char ch[];

int dp[][];

int n;

bool ok(int x,int y)

{

    if (ch[x]=='(' && ch[y]==')') return ;

    if (ch[x]=='[' && ch[y]==']') return ;

    return ;

}

int main()

{

    while(~scanf("%s",&ch))

    {

        if (ch[]=='e') break;

        n=strlen(ch);

        memset(dp,,sizeof(dp));

     //   for(int i=0;i<n;i++)

          //  for(int j=i+1;j<n;j++)  正就是不对的

        for(int i=n-;i>=;i--)

            for(int j=i+;j<n;j++)

        {

            if (ok(i,j)) dp[i][j]=max(dp[i][j],dp[i+][j-]+);

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

                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j]);

        }

        printf("%d\n",dp[][n-]);

    }

    return ;

}

巴特西

Poj 2955 brackets（区间dp）

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