POJ - 2955 Brackets 区间Dp

Brackets

Description

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 … 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

<span style="color:#000000">((()))
()()()
([]])
)[)(
([][][)
end</span>

Sample Output

<span style="color:#000000">6
6
4
0
6</span>

Source

Stanford Local 2004

区间DP基础题（点这里）：

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>
using namespace std;typedef long long ll;
const int maxn = 105;
const ll mod = 1e9+7;
const ll INF = 1e18;
const double eps = 1e-9;char s[maxn];
int dp[maxn][maxn];int main()
{while(~scanf("%s",s+1)&&s[1]!='e'){int n=strlen(s+1);memset(dp,0,sizeof(dp));for(int len=2;len<=n;len++)for(int i=1;i<=n;i++){int j=i+len-1;if(j>n) break;if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')){dp[i][j]=dp[i+1][j-1]+2;}for(int k=i;k<j;k++)dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);}printf("%d\n",dp[1][n]);}return 0;
}