DP poj 2955 Brackets

题目地址链接：http://poj.org/problem?id=2955

题目：

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 ((())) ()()() ([]]) )[)( ([][][) end Sample Output 6 6 4 0 6 Source Stanford Local 2004

代码：

#include
#include
#define maxn 110
int dp[maxn][maxn];
int main()
{
	int i,j,k,t;
	char s[maxn];
	while(gets(s)!=NULL)
	{
		if(!strcmp(s,"end"))break;
		memset(dp,0,sizeof(dp));
		int len=strlen(s);
		for(i=0;i(dp[j+1][k-1]+2)?dp[j][k]:(dp[j+1][k-1]+2);
				for(t=j;t(dp[j][t]+dp[t][k])?dp[j][k]:(dp[j][t]+dp[t][k]);
			}
		}
		printf("%d\n",dp[0][len-1]);
	}
	return 0;
}