Brackets（poj-2955）（区间dp）

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8762 Accepted: 4690

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

Source

Stanford Local 2004

#include<algorithm>#include<string.h>#include<stdio.h>#include<map>using namespace std;int dp[500][500];int cmp(char a,char b){    if(a=='['&&b==']')        return 1;    else if(a=='('&&b==')')        return 1;    else        return 0;}int main(){    char s[2000];    while(scanf("%s",s)!=EOF)    {        if(strcmp(s,"end")==0)            return 0;        memset(dp,0,sizeof(dp));        int n=strlen(s);        for(int i=0;i<n-1;i++)        {            if(cmp(s[i],s[i+1]))            {                dp[i][i+1]=2;            }        }        for(int i=2;i<n;i++)        {            for(int j=0;j<n-i;j++)            {                int k=j+i;                dp[j][k]=0;                if(cmp(s[j],s[k]))                {                    dp[j][k]=dp[j+1][k-1]+2;                }                 for(int v=j;v<k;v++)                 {                     dp[j][k]=max(dp[j][v]+dp[v+1][k],dp[j][k]);                 }            }        }        printf("%d\n",dp[0][n-1]);    }}