标签：sequence oid ons nbsp ace 个人 做了 pre sts
个人心得：今天就做了这些区间DP，这一题开始想用最长子序列那些套路的，后面发现不满足无后效性的问题，即（，）的配对
对结果有一定的影响，后面想着就用上一题的思想就慢慢的从小一步一步递增，后面想着越来越大时很多重复，应该要进行分割，
后面想想又不对，就去看题解了，没想到就是分割，还是动手能力太差，还有思维不够。
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。动态规划真的是一脸懵逼样，多思考，多瞎想吧，呼~
We give the following inductive definition of 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
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 }
标签：sequence oid ons nbsp ace 个人 做了 pre sts
原文地址：http://www.cnblogs.com/blvt/p/7371994.html
