HOJ 1936&POJ 2955 Brackets(区间DP)

Brackets
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Source : Stanford ACM Programming Contest 2004
Time limit : 1 sec Memory limit : 32 M
Submitted : 188, Accepted : 113
5.1 Description
We give the following inductive definition of a “regular brackets” sequence:
• the empty sequence is a regular brackets sequence,
• if s is a regularbrackets 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.
5.2 Example
Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].
5.3 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. For example:
((()))
()()()
([]])
)[)(
([][][)
end
5.4 Output
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line. For example:
6
6
4
0
6

一道简单的区间DP题目

关于区间DP,可以参照这个博客
http://blog.csdn.net/dacc123/article/details/50885903

#include <iostream>#include <string.h>#include <stdlib.h>#include <math.h>#include <algorithm>using namespace std;char a[105];int dp[105][105];int main(){    while(scanf("%s",a+1)!=EOF)    {        if(a[1]=='e')            break;        memset(dp,0,sizeof(dp));        int n=strlen(a+1);        for(int len=1;len<n;len++)        {            for(int i=1;i+len<=n;i++)            {                int j=i+len;                if((a[i]=='('&&a[j]==')')||(a[i]=='['&&a[j]==']'))                    dp[i][j]=dp[i+1][j-1]+2;                else                    dp[i][j]=dp[i+1][j-1];                for(int k=i;k<j;k++)                {                    if(dp[i][j]<dp[i][k]+dp[k+1][j])                        dp[i][j]=dp[i][k]+dp[k+1][j];                }            }        }        printf("%d\n",dp[1][n]);    }    return 0;}