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

6
6
4
0
6

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[105];
int dp[105][105];
int n;
bool ok(int x,int y)
{
    if (ch[x]=='(' && ch[y]==')') return 1;
    if (ch[x]=='[' && ch[y]==']') return 1;
    return 0;
}

int main()
{
    while(~scanf("%s",&ch))
    {
        if (ch[0]=='e') break;
        n=strlen(ch);
        memset(dp,0,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-1;i>=0;i--)
            for(int j=i+1;j<n;j++)
        {
            if (ok(i,j)) dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
            for(int k=i;k<=j;k++)
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j]);

        }
        printf("%d
",dp[0][n-1]);

    }
    return 0;
}