POJ 2955 Brackets

传送门@百度

Brackets

Time Limit: 1000MS		Memory Limit: 65536K

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 … 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 [([])].

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

继续刷水，和上一题差不多,dp[i][j]=max(dp[i+1][j-1]+1//i,j匹配,dp[i][k]+dp[k+1][j]);

#include<set>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 110;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)
#define Down(i,r,l) for(int i=r;i>=l;i--)

char s[N];
int dp[N][N],n;
//dp[i][j]=max{dp[i+1][j-1]+1,dp[i][k]+dp[k+1][j]}

bool match(char A,char B){
    if(A=='(') return (B==')');
    if(A=='[') return (B==']');
    return false;
}

void DP(){
    memset(dp,0,sizeof(dp));
    n=strlen(s+1);
    Down(i,n-1,1)
      Rep(j,i+1,n){
          if(match(s[i],s[j])) dp[i][j]=dp[i+1][j-1]+1;
          Rep(k,i,j-1)
            dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
      }
    cout<<dp[1][n]*2<<endl;
}

int main(){
    while(scanf("%s",s+1),strcmp(s+1,"end")){
        DP();
    }
    return 0;
}