POJ 2955 （区间dp,划分区间求解）

Description

We give the following inductive definition of 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
66406
分析：划分区间求解，状态方程dp(i,j)=max(dp(i,k)+dp(k+1,j)); dp(i,j)表示从i到j的最大长度
AC代码;
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100+10;int dp[maxn][maxn];char s[maxn];int main(){	while(scanf("%s",s+1)==1){		if(s[1]=='e')break;				int l=strlen(s+1);		memset(dp,0,sizeof(dp));		for(int len=2;len<=l;len++){  //长度 			for(int i=1;i<=l-len+1;i++) //起点 			{			 int j=i+len-1;  //终点 			 if(s[i]=='(' && s[j]==')' || s[i]=='[' && s[j]==']')			 dp[i][j]=dp[i+1][j-1]+2;			 else dp[i][j]=dp[i][j-1];			 			 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][l]);	}	return 0;}