POJ 2955-Brackets（括号匹配-区间DP）

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5484		Accepted: 2946

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

题目意思：

最大括号匹配数。

解题思路：

From

定义dp [ i ] [ j ] 为串中第 i 个到第 j 个括号的最大匹配数目

那么我们假如知道了 i 到 j 区间的最大匹配，那么i+1到 j+1区间的是不是就可以很简单的得到。

那么 假如第 i 个和第 j 个是一对匹配的括号那么 dp [ i ] [ j ] = dp [ i+1 ] [ j-1 ] + 2 ;

那么我们只需要从小到大枚举所有 i 和 j 中间的括号数目，然后满足匹配就用上面式子dp，然后每次更新dp [ i ] [ j ]为最大值即可。

更新最大值的方法是枚举 i 和 j 的中间值，然后让  dp[ i ] [ j ] = max ( dp [ i ] [ j ] , dp [ i ] [ f ] + dp [ f+1 ] [ j ] ) 。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 202
#define INF 999999
int dp[MAXN][MAXN];

int main()
{
string s;
while(cin>>s)
{
if(s=="end")  break;
memset(dp,0,sizeof(dp));
int i,j,k,f,len=s.size();
for(i=1; i<len; i++)
for(j=0,k=i; k<len; j++,k++)
{
if((s[j]=='('&&s[k]==')')||(s[j]=='['&&s[k]==']'))
dp[j][k]=dp[j+1][k-1]+2;
for(f=j; f<k; f++)
dp[j][k]=max(dp[j][k],dp[j][f]+dp[f+1][k]);
}
//len-dp[0][len-1]表示需要补充多少括号才能都匹配成功
cout<<dp[0][len-1]<<endl;
}
return 0;
}