Cf1015f bracket substring (DP + trie diagram)

Use the number of solutions for all valid sequences to subtract the number of valid sequences that do not contain the subsequences required in the question

The latter uses the AC automatic machine to maintain the DP.

 1 #include<queue> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define mod 1000000007 6 using namespace std; 7 int n,m,cnt,tot,minu,ans; 8 char son[205]; 9 int al[205][105];10 int dp[205][205][105];11 int ed[205];12 int tr[205][2];13 int fail[205];14 void build(char b[]){15     int now=0;16     int len=strlen(b+1);17     for(int i=1;i<=len;i++){18         int k=-1;19         if(b[i]==‘(‘)k=1;20         else k=0;21         if(!tr[now][k])tr[now][k]=++cnt;22         now=tr[now][k];23     }24     ed[now]=1;25 }26 void getfail(){27     queue<int>que;28     int now=0;29     if(tr[0][0])que.push(tr[0][0]);30     if(tr[0][1])que.push(tr[0][1]);31     while(!que.empty()){32         int s=que.front();33         que.pop();34         if(tr[s][0]){35             fail[tr[s][0]]=tr[fail[s]][0];36             que.push(tr[s][0]);37         }else{38             tr[s][0]=tr[fail[s]][0];39         }40         if(tr[s][1]){41             fail[tr[s][1]]=tr[fail[s]][1];42             que.push(tr[s][1]);43         }else{44             tr[s][1]=tr[fail[s]][1];45         }46     }47 }48 int main(){49     scanf("%d",&n);50     scanf("%s",son+1);51     build(son);52     getfail();53     al[1][1]=1;54     for(int i=2;i<=2*n;i++){55         for(int j=0;j<=n;j++){56             (al[i][j]+=al[i-1][j+1])%=mod;57             if(j)(al[i][j]+=al[i-1][j-1])%=mod;58         }59     }60     tot=al[2*n][0];61     dp[0][0][0]=1;62     for(int i=1;i<=2*n;i++){63         for(int j=0;j<=cnt;j++){64             for(int k=0;k<=n;k++){65                 if(!ed[tr[j][0]]){66                     (dp[i][tr[j][0]][k]+=dp[i-1][j][k+1])%=mod;67                 }68                 if(!ed[tr[j][1]]&&k){69                     (dp[i][tr[j][1]][k]+=dp[i-1][j][k-1])%=mod;70                 }71             }72         }73     }74     for(int i=0;i<cnt;i++){75         (minu+=dp[2*n][i][0])%=mod;76     }77     ans=((tot-minu)%mod+mod)%mod;78     printf("%d\n",ans);79     return 0;80 }

 

Cf1015f bracket substring (DP + trie diagram)