Brackets

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Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7728 Accepted: 4098

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

/*分类：区间DP思路：先求出需要完成括号匹配需要添加的括号如果为0，说明已经匹配的为len否则为len-已经匹配的数目 */#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;char ss[102];int main(){while(scanf("%s",ss)&&strcmp(ss,"end")!=0){int len=strlen(ss);int dp[102][102]={0};for(int i=0;i<len;i++)dp[i][i]=1;for(int i=1;i<len;i++)//区间长度 {for(int j=0;j<len-i;j++){//开始位置 int k=j+i;//结束位置 dp[j][k]=0x3f3f3f; if((ss[j]=='('&&ss[k]==')')||(ss[j]=='['&&ss[k]==']'))dp[j][k]=min(dp[j][k],dp[j+1][k-1]);for(int t=j;t<k;t++){dp[j][k]=min(dp[j][k],dp[j][t]+dp[t+1][k]);}}} if(dp[0][len-1]==0)printf("%d\n",len);else printf("%d\n",len-dp[0][len-1]);} }//思路二：直接求匹配的数目

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;char ss[102];bool  check(char c,char d){if(c=='('&&d==')')return true;if(c=='['&&d==']')return true;elsereturn false;} int main(){while(scanf("%s",ss)&&strcmp(ss,"end")!=0){int len=strlen(ss);int dp[102][102]={0};for(int i=0;i<len;i++){dp[i][i]=0;//初始化为0，匹配的数目为0 if(check(ss[i],ss[i+1]))dp[i][i+1]=2;// 已经匹配的加2 else dp[i][i+1]=0;//初始化为0，匹配的数目为0 }for(int i=1;i<len;i++)//区间长度 {for(int j=0;j<len-i;j++){//开始位置 int k=j+i;//结束位置 dp[j][k]=0; //初始化为0，匹配的数目为0 if((ss[j]=='('&&ss[k]==')')||(ss[j]=='['&&ss[k]==']'))dp[j][k]=max(dp[j][k],dp[j+1][k-1]+2);for(int t=j;t<k;t++){dp[j][k]=max(dp[j][k],dp[j][t]+dp[t+1][k]);}}}printf("%d\n",dp[0][len-1]);}}


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