Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 2999 | Accepted: 1536 |
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
Source
代碼:
#include <iostream>
#include <string.h>
#include <math.h>
#include <stdio.h>
using namespace std;
#define M 110
int dp[M][M];
char str[M];
int math(int x,int y) //判斷是否配對。
{
return str[x]=='(' && str[y]==')'||
str[x]=='{' && str[y]=='}'||
str[x]=='[' && str[y]==']';
}
int DFS(int s,int e)
{
int i,ret,t;
if(e<s) return 0; //對各種特殊情況進行考慮。
if(e==s) return dp[s][e]=0;
if(e-s==1) return dp[s][e]=math(s,e);
if(dp[s][e]!=-1) return dp[s][e]; //如果這個狀態已經計算過。
ret=DFS(s+1,e); //姿勢不對,還是看別人的,不懂。
for(i=s+1;i<=e;i++) //找能配對的括號。
if(math(s,i))
{
t=DFS(s+1,i-1)+DFS(i+1,e)+1;
if(t>ret) ret=t;
}
return dp[s][e]=ret;
}
int main()
{
int i,j,k,n;
while(scanf("%s",str))
{
if(str[0]=='e') break;
n=strlen(str);
memset(dp,-1,sizeof(dp));
DFS(0,n-1);
printf("%d\n",dp[0][n-1]*2);
}
return 0;
}