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
第二次写这道题了，对区间dp更了解了，通过对区间设置分段点进行区间dp,将两个区间的最大值可以合在一起
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define ls (k<<1)
#define rs (k<<1|1)
#define pb push_back
#define mid ((l+r)>>1)
using namespace std;
const int p=1e4+7;
const int mod=1e9+7;
const int maxn=200;
typedef long long ll;
const int inf=0x3f3f3f3f;
int dp[maxn][maxn];
char s[maxn];
void solve(){
while(scanf("%s",s+1)!=EOF){
if(s[1]=='e') break;
memset(dp,0,sizeof(dp));
int len=strlen(s+1);
for(int i=1;i<=len;i++){
//枚举字符串长度
for(int j=1;j+i<=len+1;j++){
//枚举端点
int k=i+j-1;//
if(s[j]=='('&&s[k]==')')
dp[j][k]=dp[j+1][k-1]+2;//(p)
else if(s[j]=='['&&s[k]==']')
dp[j][k]=dp[j+1][k-1]+2;
for(int x=j;x<k;x++)//枚举区域中的分端点
dp[j][k]=max(dp[j][k],dp[j][x]+dp[x+1][k]);//()()()
}
}
cout<<dp[1][len]<<endl;
}
}
int main(){
ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
solve();
return 0;
}