個人心得：今天就做了這些區間DP，這一題開始想用最長子序列那些套路的，后面發現不滿足無后效性的問題，即（，）的配對

對結果有一定的影響，后面想著就用上一題的思想就慢慢的從小一步一步遞增，后面想著越來越大時很多重復，應該要進行分割，

后面想想又不對，就去看題解了，沒想到就是分割，還是動手能力太差，還有思維不夠。

1 for(int j=0;j+i<ch.size();j++) 2 { 3 if(check(j,j+i)) 4 dp[j][j+i]=dp[j+1][j+i-1]+2; 5 for(int m=j;m<=j+i;m++) 6 dp[j][j+i]=max(dp[j][j+i],dp[j][m]+dp[m+1][j+i]); 7 }

分割并一次求最大值。動態規劃真的是一臉懵逼樣，多思考，多瞎想吧，呼~

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

6 6 4 0 6 1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<iomanip> 6 #include<string> 7 #include<algorithm> 8 using namespace std; 9 int money[205]; 10 int dp[205][205]; 11 string ch; 12 const int inf=999999; 13 int check(int i,int j){ 14 if((ch[i]=='('&&ch[j]==')')||(ch[i]=='['&&ch[j]==']')) 15 return 1; 16 return 0; 17 } 18 void init(){ 19 for(int i=0;i<ch.size();i++) 20 for(int j=0;j<ch.size();j++) 21 dp[i][j]=0; 22 } 23 int main(){ 24 int n,m; 25 while(getline(cin,ch,'\n')){ 26 if(ch=="end") break; 27 init(); 28 for(int k=0;k<ch.size()-1;k++) 29 if(check(k,k+1)) 30 dp[k][k+1]=2; 31 else 32 dp[k][k+1]=0; 33 for(int i=2;i<ch.size();i++) 34 { 35 for(int j=0;j+i<ch.size();j++) 36 { 37 if(check(j,j+i)) 38 dp[j][j+i]=dp[j+1][j+i-1]+2; 39 for(int m=j;m<=j+i;m++) 40 dp[j][j+i]=max(dp[j][j+i],dp[j][m]+dp[m+1][j+i]); 41 } 42 43 } 44 cout<<dp[0][ch.size()-1]<<endl; 45 } 46 return 0; 47 }

?

轉載于:https://www.cnblogs.com/blvt/p/7371994.html

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