POJ 2955 Brackets（括号最大匹配，区间DP）

原题地址

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6711		Accepted: 3612

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, …, imwhere 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

求能匹配上的括号的总数

用dp[ i ][ j ]代表区间[ i , j ]内能匹配上的括号的总数

AC代码：
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>

using namespace std;

char ch[200];
int dp[106][106];
int main()
{
while(~scanf("%s",&ch))
{
if(strcmp(ch,"end")==0)
{
break;
}
int i,j,k,l;
memset(dp,0,sizeof(dp));
l=strlen(ch);
for(i=1;i<l;i++)
{
for(j=0,k=i;k<l;j++,k++)
{
if(ch[j]=='('&&ch[k]==')'||ch[j]=='['&&ch[k]==']')
dp[j][k]=dp[j+1][k-1]+2;
for(int p=j;p<k;p++)
{
if(dp[j][k]<dp[j][p]+dp[p+1][k])
{
dp[j][k]=dp[j][p]+dp[p+1][k];
}
}
}
}
printf("%d\n",dp[0][l-1]);
}
return 0;
}