Brackets, adobebrackets
Brackets
Time Limit:1000 MS |
|
Memory Limit:65536 K |
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,
- IfSIs a regular brackets sequence, then (S) And [S] Are regular brackets sequences, and
- IfAAndBAre regular brackets sequences, thenABIs 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 charactersA1A2...An, Your goal is to find the length of the longest regular brackets sequence that is a subsequenceS. That is, you wish to find the largestMSuch that for indicesI1,I2 ,...,ImWhere 1 ≤I1 <I2 <... <Im≤N,Ai1Ai2...AimIs 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, intrusive. The end-of-file is marked by a line containing the word "end" and shocould not be processed.
Output
For each input case, the program shocould print the length of the longest possible regular brackets subsequence on a single line.
Sample Input
((()))()()()([]]))[)(([][][)end
Sample Output
66406
Source
Stanford Local 2004
Interval DP according to rules
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 }