"蔚来杯"2022牛客暑期多校训练营2 KLink with Bracket Sequence I

问题描述

Link has a bracket sequence a of length n, which is a subsequence of a valid bracket sequence b of length m.

Link doesn't remember b, so he wonders the number of possible sequences b.

A bracket sequence is valid if it satisfies any of the following conditions:

• Its length is 0.

• It can be represented as (A), where A is a valid bracket sequences.

• It can be represented as AB, where A and B are both valid bracket sequence.

A sequence a is a subsequence of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements.

输入格式

Each test contains multiple test cases. The first line contains the number of test cases T (1≤T≤100). Description of the test cases follows.

The first line contains two integers n,m (1≤n≤m≤200).

The second line contains a bracket sequence s of length n.

It is guaranteed that the sum of m over all test cases does not exceed 10³.

输出格式

For each test cases, output the number of possible sequences b modulo 10⁹+7.

Note that the answer maybe 0 due to Link's poor memory.

样例输入

3
2 2
()
2 4
)(
2 4
()

样例输出

1
1
2

题解

Link有一个长度为n的括号序列a，它是长度为m的合法括号序列m的子序列

Link不记得b，所以他想知道可能的b的个数

一个合法的括号序列应满足以下条件：

• 序列长度为0
• 若A是合法序列，则（A）也是合法序列
• 若A和B是合法序列，则AB是合法序列

  ---

设f[i][j][k]表示b的长度为i，a的前j个是b的子序列，且左括号个数比右括号多k（即有k的单独的左括号）的方案数

为什么左括号个数不能少于右括号

因为我们构造b时是通过在字符串末尾添加字符构造的，如果左括号个数大于右括号，可以通过后期加右括号使最终序列合法，如果右括号多于左括号，要使最终序列合法需要在前面添加左括号，但是我们构造b时只能在后面添加字符，所以若右括号多于左括号就不可能构成合法序列

显然，f[0][0][0]=1

设当前已构造出长度为i的b，对于b的第i+1个字符

第一种情况：左括号

  当a[i+1]是左括号时，直接将a[i+1]接在b末尾，f[i+1][j+1][k+1]+=f[i][j][k]

  当a[i+1]是右括号时，不使用a[i+1]，在b末尾额外加一个左括号f[i+1][j][k+1]+=f[i][j][k]

第二种情况：右括号

  每加一个右括号，前面必须有单独的左括号与其配对，即左括号的个数要严格大于右括号

  而由于左括号被配对掉一个，左右括号个数差变为k-1

  当a[i+1]是左括号时，不使用a[i+1]，直接在b末尾加一个右括号，f[i+1][j][k-1]+=f[i][j][k]

  当a[i+1]是右括号时，将a[i+1]接在b末尾，f[i+1][j+1][k-1]+=f[i][j][k]

#include <cstring>
#include <cstdio>
const int mo = 1e9+7;
int T,n,m,f[205][205][205];
char s[205];
void add(int &x,int y)
{
    x=(x+y)%mo;
    return;
}
int main()
{
    int i,j,k;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d%s",&n,&m,s+1); 
        memset(f,0,sizeof(f));
        f[0][0][0]=1; 
    for (i=0;i<m;i++) 
      for (j=0;j<=n;j++)
        for (k=0;k<=i;k++)
        {
            add(f[i+1][j+(s[j+1]=='(')][k+1],f[i][j][k]);
            if (k) add(f[i+1][j+(s[j+1]==')')][k-1],f[i][j][k]);
        }
        printf("%d\n",f[m][n][0]);
    }
    return 0;
}