Brackets
2018-06-17 22:12:29来源:未知 阅读 ()
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8085 | Accepted: 4299 |
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
6 6 4 0 6
Source
根据规则来进行区间DP
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #define lli long long int 6 using namespace std; 7 const int MAXN=1001; 8 const int maxn=0x7fffff; 9 void read(int &n) 10 { 11 char c='+';int x=0;bool flag=0; 12 while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;} 13 while(c>='0'&&c<='9') 14 x=(x<<1)+(x<<3)+c-48,c=getchar(); 15 flag==1?n=-x:n=x; 16 } 17 char s[MAXN]; 18 int dp[MAXN][MAXN]; 19 int main() 20 { 21 while(scanf("%s",s)) 22 { 23 if(s[0]=='e') 24 break; 25 memset(dp,0,sizeof(dp)); 26 int l=strlen(s); 27 for(int i=l;i>=0;i--) 28 for(int j=i;j<l;j++) 29 { 30 //dp[i][j]=max(dp[i+1][j],dp[i][j-1]); 31 32 if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')) 33 dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2); 34 35 for(int k=i;k<j;k++) 36 dp[i][j]=max(dp[i][k]+dp[k+1][j],dp[i][j]); 37 38 } 39 printf("%d\n",dp[0][l-1]); 40 } 41 return 0; 42 }
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