Portal @ Baidu
Brackets
Time limit:1000 ms |
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Memory limit:65536 K |
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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
Continue refreshing, similar to the previous question, DP [I] [J] = max (DP [I + 1] [J-1] + 1 // I, j match, DP [I] [k] + dp [k + 1] [J]);
1 #include<set> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 const int N = 110; 9 #define For(i,n) for(int i=1;i<=n;i++)10 #define Rep(i,l,r) for(int i=l;i<=r;i++)11 #define Down(i,r,l) for(int i=r;i>=l;i--)12 13 char s[N];14 int dp[N][N],n;15 //dp[i][j]=max{dp[i+1][j-1]+1,dp[i][k]+dp[k+1][j]}16 17 bool match(char A,char B){18 if(A==‘(‘) return (B==‘)‘);19 if(A==‘[‘) return (B==‘]‘);20 return false;21 }22 23 void DP(){24 memset(dp,0,sizeof(dp));25 n=strlen(s+1);26 Down(i,n-1,1)27 Rep(j,i+1,n){28 if(match(s[i],s[j])) dp[i][j]=dp[i+1][j-1]+1;29 Rep(k,i,j-1)30 dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);31 }32 cout<<dp[1][n]*2<<endl;33 }34 35 int main(){36 while(scanf("%s",s+1),strcmp(s+1,"end")){37 DP();38 }39 return 0;40 }
Codes
Poj 2955 brackets