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Description
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
求括号能够正确匹配的括号个数
dp（i,j）表示区间ij内的正确括号个数
dp（i,j）=max｛dp（i+1，j-1）+2 （仅当外层括号匹配），dp（i，k）+dp（k，j）（进行内层括号检查）｝
#include"iostream" #include"cstring" using namespace std; int dp[110][110]; string a; void Work() { int len=a.length(); //cout<<len<<endl; memset(dp,0,sizeof(dp)); int ans=0; for(int i=0;i<len;i++) { for(int j=0,k=i;k<len;k++,j++) { if((a[j]==‘(‘&&a[k]==‘)‘)||(a[j]==‘[‘&&a[k]==‘]‘)) dp[j][k]=dp[j+1][k-1]+2; for(int t=j+1;t<k;t++) if(dp[j][t]+dp[t][k]>dp[j][k]) dp[j][k]=dp[j][t]+dp[t][k]; if(dp[j][k]>ans) ans=dp[j][k]; } } cout<<ans<<endl; } int main() { while(cin>>a) { if(a=="end") break; Work(); } return 0; }
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原文地址：http://www.cnblogs.com/zsyacm666666/p/4725660.html
