POJ 2955 Brackets (区间DP)

Brackets

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 6420 Accepted: 3436

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

Stanford Local 2004

给出一个串，请取出其中的一些字符组成的新串括号必须是两两匹配的。可以不连续地取，但必须保持先后顺序。

#include "cstring"
#include "cstdio"
#include "iostream"
#include "string.h"
#include "algorithm"
using namespace std;
char s[1005];
int dp[1005][1005];
bool check(int left,int right)
{
if(s[left]=='('&&s[right]==')')
return true;
if(s[left]=='['&&s[right]==']')
return true;
return false;
}
int main()
{
while(scanf("%s",s)&&strcmp(s,"end")!=0)
{
int len=strlen(s);
memset(dp,0,sizeof(dp));
//l表示区间长度。区间的长度可以从1~len
for(int l=1;l<=len;l++)
{
//枚举该长度下的所有区间
for(int i=0;i+l-1<len;i++)
{
//若区间守卫可以构成一个匹配，那么答案就+2
if(check(i,i+l-1))
dp[i][i+l-1]=dp[i+1][i+l-2]+2;
//合并区间
for(int j=i;j<i+l-1;j++)
dp[i][i+l-1]=max(dp[i][i+l-1],dp[i][j]+dp[j+1][i+l-1]);
}
}
printf("%d\n",dp[0][len-1]);
}

}