題目

農夫約翰的奶牛們排成了一條直線，每隻奶牛都有一個識別符號，這個識別符號是左括號或者右括號。約翰希望將這個奶牛序列分成兩個子序列，並且不改變序列中原來的字元順序。同時，要求這兩個子序列都是平衡序列。 平衡序列是這麼定義的：序列中左括號和右括號的個數是一樣的，同時這個序列的所有字首串中左括號的個數不少於右括號的個數。

例如圖1都是平衡序列。 () (()) ()(()()) 圖2都不是平衡序列。 )( ())( ((()))) 請你幫助約翰計算將原來的奶牛序列分成兩個平衡子序列的方案數。

輸入

只有一個奶牛的序列，都是由左括號和右括號組成的。

輸出

總的方案數。

樣例輸入

(())

樣例輸出

6

資料範圍限制

序列的長度是$1$到$1000$。

題解

We’ll call the two breeds A and B for convenience. Let the input string $S = s_1s_2...s_n$. We will give a dynamic programming algorithm working backwards from the end of the string.

Let $f(i,A_{open}, B_{open})$ be the number of ways to assign $s_{}$

i...sns_i...s_n to breeds such that the resulting two parentheses-strings are balanced, given that we have $A_{open}$ unmatched left-parenthesis of type A and B_open unmatched left-parentheses of type B. If $S[i]=='('$, then $f(i,A_{open},B_{open})=f(i+1,$

Aopen+1,Bopen)+f(i+1,Aopen,Bopen+1)f(i,A_{open}, B_{open}) = f(i+1, A_{open}+1, B_{open}) + f(i+1, A_{open}, B_{open} + 1), since we can assign the parenthesis to breed A or breed B. If $S[i]==')'$, then we can assign the parenthesis to breed A as long as $A_{open} > 0$, and to B as long as $B_{open} > 0$.

The base case is i=n, in which case we processed the whole string without violating any invariants. As the total number of ')'s equals to the total number of $'('$s, we wil end up with two balanced strings of parentheses. Therefore we can start with so $f(n, 0, 0) = 1$.

We have $0 <= i <= n, 0 <= A_{open} <= n, 0 <= B_{open} <= n$, so the number of states is $O(n^3)$, and there is $O(1)$ non-recursive overhead for each state, so this leads to an $O(n^3)$solution.

Unfortunately, $O(n^3)$ isn’t fast enough with $n=1,000$. We can do better by noticing that $B_{open}$ is uniquely determined by i and $A_{open} (because A_{open} + B_{open}$ sums to the number of unmatched left-parentheses in $s_1...s_{i-1})$. So it suffices to keep track of $(i, A_{open})$, which gives $O(n^2) states \ and\ an O(n^2)$ solution. This is fast enough.

Code

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
using namespace std;
#define MOD 2012
#define MAXN 1010
int a[MAXN];
int main() 
{
    freopen("bbreeds.in","r",stdin);
    freopen("bbreeds.out","w",stdout);
    int l = a[1] = 1;
    for(int ch = cin.get();l > 0 && ch == '(' || ch == ')';ch = cin.get()) 
    {
        int dir = ch == '(' ? 1 : -1;
        l += dir;
        for(int j = dir < 0 ? 1 : l;1 <= j && j <= l; j -= dir) 
        {
            a[j] += a[j - dir];
            if(a[j] >= MOD) 
                a[j] -= MOD;
        }
    a[l + 1] = 0;
  }
  cout << (l == 1 ? a[1] : 0) << endl;
}