Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 17416 | Accepted: 9016 |
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 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.
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
#include<stdio.h>
#include<string.h>
#define maxn 110
int dp[maxn][maxn];
int max(int x,int y){
return x>y?x:y;
}
int main(){
char s[maxn];
while(scanf("%s",s)!=EOF){
int i,j,k,len,n;
if(s[0]=='e') break;
memset(dp,0,sizeof(dp));
n=strlen(s);
for(len=1;len<=n;len++){
for(i=0;i<n;i++){
j=len+i-1;
//if(s[i]==')'||s[i]==']')
dp[i][j]=dp[i+1][j]; //第i个在这段区间没有匹配
//else
for(k=i+1;k<=j;k++){ //第i个与第k个位置匹配上时,状态如下
if((s[k]==')'&&s[i]=='(')||(s[k]==']'&&s[i]=='['))
dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
}
}
}
printf("%d\n",dp[0][n-1]);
}
return 0;
}