HOJ 1936&POJ 2955 Brackets(区间DP)

Brackets My Tags (Edit) 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

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>

using namespace std;
char a;
int dp;
int main()
{
while(scanf("%s",a+1)!=EOF)
{
if(a=='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[n]);
}
return 0;
}

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