A - Brackets

A - Brackets
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d
& %I64u
Submit Status

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

题意是给你一串括弧符号，要判断一共可以有多少个配对的，只有(),[],这两种配对方法。

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int dp[110][110];
char s[110];
bool match(char a,char b)
{
if(b-a==1||b-a==2)
return true;
else return false;
}
int main()
{
while(~scanf("%s",s)&&s[0]!='e')
{
int len=strlen(s);
memset(dp,0,sizeof(dp));
for(int i=0;i<len;i++)
{
if(match(s[i],s[i+1]))
dp[i][i+1]=2; //一个肯定是不能配对成功的，所以直接预处理长度为2 的，如果能配对则为2，否则为0
}
for(int k=3;k<=len;k++) //长度为1和2 的都处理过了，直接从长度为3的开始计算
{
for(int i=0;i+k-1<len;i++){
if(match(s[i],s[i+k-1]))
dp[i][i+k-1]=dp[i+1][i+k-2]+2; //如果长度为k的区间两端的括弧能够配对，则等于中间的配对数加上2
for(int j=i;j<i+k-1;j++) //从i到长度为k的最后一位，一直枚举断点
{
dp[i][i+k-1]=max(dp[i][i+k-1],dp[i][j]+dp[j+1][i+k-1]); //取某个断点能组成的最大配对数
}
}
}
printf("%d\n",dp[0][len-1]);
}
return 0;
}