POJ - 2955 Brackets (區間DP)
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
Given the initial sequence ([([]])]
[([])]
.
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
思路:從後往回找,如果找到一個‘)’或‘]’與當前的'('或'['對應,則加2。dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2)。
#include<algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
char a[110],mp[200];
int dp[110][110];
int main()
{
mp['(']=')';
mp['[']=']';
while(~scanf("%s",a),strcmp(a,"end"))
{
int len=strlen(a);
memset(dp,0,sizeof(dp));
for(int i=len-1;i>=0;i--)
{
for(int j=i+1;j<len;j++)
{
dp[i][j]=dp[i+1][j];
for(int k=i;k<=j;k++)
{
if(mp[a[i]]==a[k])
{
dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
}
}
}
}
printf("%d\n",dp[0][len-1]);
}
}