Codeforces Round #501 (Div. 3) F. Bracket Substring

https://codeforces.com/problemset/problem/1015/F

dp

'求包含某个子串的个数' 类型

kmp

    ///it is advised that all character begins at index 1

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define ll long long

const double eps=1e-8;
const ll inf=1e9;
const ll mod=1e9+7;
const int maxn=2e2+10;

char sstr[maxn];

int str[maxn],pre[maxn],nex[maxn][2];
ll f[maxn][maxn][maxn][2];

int main()
{
    int n,nn,len,i,j,k,l,m,jj,lll;
    scanf("%d",&n);
    nn=n<<1;
    ///it is advised that all character begins at index 1
    scanf("%s",sstr+1);
    len=strlen(sstr+1);
    for (i=1;i<=len;i++)
        if (sstr[i]=='(')
            str[i]=0;
        else
            str[i]=1;

    ///kmp makes O(1) visit
    j=0;
    pre[1]=0;
    for (i=2;i<=len;i++)
    {
        while (j!=0 && str[j+1]!=str[i])
            j=pre[j];
        if (str[j+1]==str[i])
            j++;
        pre[i]=j;
    }

    for (i=0;i<=len-1;i++)
        for (j=0;j<=1;j++)
            if (str[i+1]==j)
                nex[i][j]=i+1;
            else
                nex[i][j]=nex[pre[i]][j];

    f[0][0][0][0]=1;
    ///pos ith
    for (i=1;i<=nn;i++)
        ///j '('
        for (j=0;j<=min(i-1,n);j++)
            ///current char
            for (k=0;k<=1;k++)
            {
                jj=j+1+(-2)*(k==1);
                if (jj==-1)
                    continue;
                (f[i][jj][0][1]+=f[i-1][j][0][1])%=mod;
                ///the lth str
                for (l=0;l<=min(len-1,i-1);l++)
                {
                    lll=nex[l][k];
                    if (lll==len)
                        (f[i][jj][0][1]+=f[i-1][j][l][0])%=mod;
                    else
                        (f[i][jj][lll][0]+=f[i-1][j][l][0])%=mod;
                }
            }
    printf("%lld",f[nn][0][0][1]);
    return 0;
}