poj 2955 Brackets（区间dp）

题目链接

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3470		Accepted: 1792

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

题意：给一个只包含括号的串，问该串的最长的合法括号匹配的子序列的长度？

题解：先考虑第一个括号，有两种操作：删除这个括号；将该括号与另一个括号匹配。

用dp[l][r]表示区间[l,r]的最长合法子序列的长度，转移就是：

1，删除第一个括号

dp[l][r]=dp[l+1][r]

2，将第一个括号和第i个括号匹配

dp[l][r]=min(dp[l+1][i-1]+dp[i+1][r]+2);

代码如下：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<string>
#include<math.h>
#define nn 110
#define inff 0x7fffffff
#define eps 1e-8
typedef long long LL;
const LL inf64=LL(inff)*inff;
using namespace std;
int n;
char s[nn];
int dp[nn][nn];
int lj[nn][nn];
bool check(int l,int r)
{
if(s[l]=='('&&s[r]==')')
return true;
if(s[l]=='['&&s[r]==']')
return true;
return false;
}
int dfs(int l,int r)
{
if(dp[l][r]!=-1)
return dp[l][r];
if(l>r)
return 0;
if(l==r)
return dp[l][r]=0;
dp[l][r]=0;
int i;
for(i=l+1;i<=r;i++)
{
if(check(l,i))
dp[l][r]=max(dp[l][r],dfs(l+1,i-1)+dfs(i+1,r)+2);
}
dp[l][r]=max(dp[l][r],dfs(l+1,r));
return dp[l][r];
}
int main()
{
int i;
while(scanf("%s",s)!=EOF)
{
if(strcmp(s,"end")==0)
break;
int len=strlen(s);
memset(dp,-1,sizeof(dp));
printf("%d\n",dfs(0,len-1));
}
return 0;
}