POJ 题目2955 Brackets（区间dp）

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3444		Accepted: 1770

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_mwhere 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

问有多少个括号匹配

ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
int dp[1010][1010];
char s[110];
int jud(int i,int j)
{
	if(s[i]=='('&&s[j]==')')
		return 1;
	if(s[i]=='['&&s[j]==']')
		return 1;
	return 0;
}
int main()
{
	//char s[110];
	while(scanf("%s",s)!=EOF)
	{
		int len,k,i,j,r;
		memset(dp,0,sizeof(dp));
		if(strcmp("end",s)==0)
			break;
		len=strlen(s);
		for(i=0;i<len-1;i++)
		{
			if(jud(i,i+1))
				dp[i][i+1]=2;
			else
				dp[i][i+1]=0;
		}
		for(k=2;k<len;k++)
		{
			for(i=0;i+k<len;i++)
			{
				r=k+i;
				if(jud(i,r))
					dp[i][r]=dp[i+1][r-1]+2;
				for(j=i;j<r;j++)//j从i开始
				{	
					dp[i][r]=max(dp[i][r],dp[i][j]+dp[j+1][r]);
				}
			}
		}
		printf("%d\n",dp[0][len-1]);
	}
}