poj--2955--Brackets（区间dp）

Brackets

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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

区间dp，每次在i--j寻找最大的匹配数，更新dp[i][j]，每一步都应该有dp[i][j]=dp[i+1][j]，因为下一步循环不一定能进入

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[10010];
int dp[1010][1010];
bool judge(char a,char b)
{
	if(a=='('&&b==')') return true;
	if(a=='['&&b==']') return true;
	return false;
}
int main()
{
	while(scanf("%s",str)!=EOF)
	{
		if(strcmp(str,"end")==0) break;
		int len=strlen(str);
		memset(dp,0,sizeof(dp));
		for(int i=len-1;i>=0;i--)
		{
			for(int j=i;j<len;j++)
			{
				dp[i][j]=dp[i+1][j];
				for(int k=i+1;k<=j;k++)
				if(judge(str[i],str[k]))
				dp[i][j]=max(dp[i][j],dp[i+1][k-1]+1+dp[k+1][j]);
			}
		}
		memset(str,'\0',sizeof(str));
		printf("%d\n",dp[0][len-1]*2);
	}
	return 0;
}