POJ 2955-Brackets

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 … a_n, 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, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

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.

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
using namespace std;
const int maxn = 100+10;
string st;
int n;
int dp[maxn][maxn];
map<char,char> mma;

int dfs(int sta,int ed){
    if(sta>=ed) return 0;
    if(dp[sta][ed] != -1) return dp[sta][ed];
    int ans = 0;
    if(mma[st[sta]]==st[ed] ) ans = dfs(sta+1,ed-1)+2;
    for(int i = sta; i < ed; i++){
            ans = max(ans,dfs(sta,i)+dfs(i+1,ed));
    }
    return dp[sta][ed] = ans;
}

int main(){
    mma[‘(‘] = ‘)‘;
    mma[‘[‘] = ‘]‘;
    while(cin >> st && st != "end"){
        memset(dp,-1,sizeof dp);
        n = st.size();
        cout<<dfs(0,n-1)<<endl;
    }
    return 0;
}