POJ 2955 （区间dp,划分区间求解）

Love The Way You Lie 2024-02-17 19:36 54阅读 0赞

Brackets

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;7285</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;3892</td> 
  </tr> 
 </tbody> 
</table>

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 *a*1*a*2 … *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 *i*1, *i*2, …, *im* where 1 ≤ *i*1 < *i*2 < … < *im* ≤ *n*, *ai*1*ai*2 … *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

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;
    }