POJ 2955 Brackets

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7607 Accepted: 4043

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

66406

找到拥有括号层数最多的以堆括号有多少个括号，()2个，([)]2个，([])]4个，([])]]]]4个，((()))6个。

区间就是第i个和第i+k之间的遍历一遍，k就代表区间大小。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char s[102];int dp[102][102];int main(){    int i,j,k,n;    while(~scanf("%s",s))    {        if(s[0]=='e')        break;        n=strlen(s);        memset(dp,0,sizeof(dp));        for(k=1;k<n;k++)        {            for(i=0;i+k<n;i++)            {                if(s[i]=='('&&s[i+k]==')'||s[i]=='['&&s[i+k]==']')                dp[i][i+k]=dp[i+1][i+k-1]+2;                for(j=i;j<i+k;j++)                {                    dp[i][i+k]=max(dp[i][i+k],dp[i][j]+dp[j+1][i+k]);                }            }        }        printf("%d\n",dp[0][n-1]);        memset(s,0,sizeof(s));    }    return 0;}