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poj 2955 Brackets（区间dp）

2024-09-19 16:22:47   219人阅读

                                                                                                Brackets
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7756 Accepted: 4114

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 … an, 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, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim 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
以前写过一道类似的题 最优矩阵乘法的那个 也是一道dp 发现这叫区间dp
dp[i][j]表示【i j]这个区间的最大匹配个数
我们可以通过求【i，k],[k+1,j]来求这个【i,j]
所以我们只要枚举这个区间找到最大的那个
转移方程 dp[i][j]=max(dp[i][k],dp[k+1][j]);
 1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cctype> 5 #include<cmath> 6 #include<cstring> 7 #include<map> 8 #include<stack> 9 #include<set>10 #include<vector>11 #include<algorithm>12 #include<string.h>13 typedef long long ll;14 typedef unsigned long long LL;15 using namespace std;16 const int INF=0x3f3f3f3f;17 const double eps=0.0000000001;18 const int N=1000+10;19 char str[N];20 int dp[N][N];21 int judge(int x,int y){22     if(str[x]==‘(‘&&str[y]==‘)‘)return 1;23     if(str[x]==‘[‘&&str[y]==‘]‘)return 1;24     return 0;25 }26 int main(){27     while(gets(str)){28         if(strcmp(str,"end")==0)break;29         memset(dp,0,sizeof(dp));30         int len=strlen(str);31         for(int t=1;t<len;t++){32             for(int i=0;i+t<len;i++){33                 int j=i+t;34                 if(judge(i,j)){35                     if(j-i==1)dp[i][j]=2;36                     else dp[i][j]=dp[i+1][j-1]+2;37                 }38                 for(int k=i;k<j;k++){39                     dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);40                 }41             }42         }43         cout<<dp[0][len-1]<<endl;44     }45 }

 

poj 2955 Brackets（区间dp）


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