poj-2955-Brackets-区间DP

poj-2955-Brackets-区间DP

 

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9014		Accepted: 4829

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 … 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

Source

Stanford Local 2004

 

 

一开始使用code2，找到总的pair数量再*2，以为就解决了。但是wrong answer，不知道本code的方法错在哪里？

使用了区间DP，方法参考自： http://www.cnblogs.com/kuangbin/archive/2013/04/29/3051402.html

 

2955

Accepted

472K

47MS

G++

858B

2017-09-16 11:12:08

 

#include <cstdio> 
#include <cstring>const int MAXN = 110; #define max(a, b) (a)>(b)?(a):(b) char ch[MAXN]; 
int dp[MAXN][MAXN]; int Solve(int l, int r){if(dp[l][r] != -1){return dp[l][r]; }if(l>=r){dp[l][r] = 0; return 0; }else if(r == l +1 ){if((ch[l]=='(' && ch[r]==')') ||  (ch[l]=='[' && ch[r]==']') ){dp[l][r] = 2; }else{dp[l][r] = 0; }return dp[l][r]; }dp[l][r] = Solve(l+1, r); for(int k=l+1; k<=r; ++k){if((ch[l]=='(' && ch[k]==')') || (ch[l]=='[' && ch[k]==']')  ){dp[l][r] = max(dp[l][r], 2+Solve(l+1, k-1) + Solve(k+1, r)); }}return dp[l][r]; 
}int main(){freopen("in.txt", "r", stdin); while(scanf("%s", ch) != EOF){getchar(); if(strcmp(ch, "end") == 0){break; }int len = strlen(ch); memset(dp, -1, sizeof(dp)); int ans = Solve(0, len-1); printf("%d\n", ans );}return 0; 
}

　　

 

采用count的方式来计算 subsequence，不知道wrong answer的地方是啥？

 

#include <cstdio>  
#include <cstdlib>  #include <cstring> const int MAXN = 12; int main(){freopen("in.txt", "r", stdin);  int ans, f1,f2, len; char ch[MAXN]; while(scanf("%s", ch) != EOF){getchar(); if(strcmp(ch, "end") == 0){break; } len = strlen(ch); ans = 0;f1 = 0; f2 = 0;  for(int i=0; i<len; ++i){if(ch[i] == '('){++f1; }else if(ch[i] == '['){++f2; }else if(ch[i] == ')'){if(f1 > 0){++ans; --f1; }}else if(ch[i] == ']'){if(f2 > 0){++ans; --f2; }}}printf("%d\n", 2*ans );} return 0; 
}