stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;



#define ll long long

#define size 100001



int a[size];

int b[size];

int c[size];

int dp[size][4];



int max_sum(int n, int p)

{

    if(n <= 0)

        return 0;

    if(n > 0 && dp[n][p] != -1)

        return dp[n][p];

    int ans = INT_MIN;

    if(p != 0)

    {

        ans = max(ans,a[n]+max_sum(n-1,0));

    }

    if(p != 1)

    {

        ans = max(ans,b[n]+max_sum(n-1,1));

    }

    if(p != 2)

    {

        ans = max(ans,c[n]+max_sum(n-1,2));

    }



    return dp[n][p] = ans;

}







int main()

{

  //OJ;

  int t = 1;

  //cin>>t;

  while(t--){

   memset(dp,-1,sizeof(dp));

  int n;

  cin>>n;



  for(int i = 1; i <= n; i++)

  {

      cin>>a[i]>>b[i]>>c[i];

  }

  cout<<max_sum(n,3)<<endl;

  }

   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.578310430868698x):
#include <bits/stdc++.h>

 

#define f(i,a,b) for( ll i = a; i < b ; i++ )

#define af(i,a,b) for( ll i = a; i >= b ; i--)

#define rep(i,a,b,k) for(ll i = a; i < b ; i+= k )

#define arep(i,a,b,k) for( ll i = a; i >= b ; i-= k)

#define ones(x) (ll) __builtin_popcount(x)

#define fs first

#define sc second

#define pb push_back

#define po pop_back

#define mp make_pair  

#define sz(a) (ll) a.size()

#define all(a) a.begin(), a.end()

#define sor(a) sort( a.begin(), a.end() )

#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL)

#define ller ios::sync_with_stdio(false);cin.tsie(nullptr);cout.tie(nullptr)

#define watch(x) cout << (#x) << " is #" << (x) <<"#" <<endl

#define PI 3.1415926535

using namespace std;

typedef long long ll;

typedef long double ld; 

typedef pair<ll,ll> ii ;

typedef vector<ll>  vi ;

typedef vector<ii> vii ;

const ll mod = 1e9+7;

const ll MAX = 1e5+5;

const ll inf = 1e18+7;

ll dp[MAX][4];

ll a[MAX],b[MAX],c[MAX];

int main() {

  fastio;

  ll n;

  cin>>n;

  f(i,0,n) cin>>a[i]>>b[i]>>c[i];

  dp[0][0]=a[0];

  dp[0][1]=b[0];

  dp[0][2]=c[0];

  f(i,1,n){

    dp[i][0] = max(dp[i-1][1]+a[i],dp[i-1][2]+a[i]);

    dp[i][1] = max(dp[i-1][0]+b[i],dp[i-1][2]+b[i]);

    dp[i][2] = max(dp[i-1][0]+c[i],dp[i-1][1]+c[i]);

  }

  ll ans = max(dp[n-1][0],max(dp[n-1][1],dp[n-1][2]));

  cout<<ans<<"\n";

  return 0; 

}
// Slower C++ solution:
#include <bits/stdc++.h>





using namespace std;



vector<vector<int>> vec(100007);

int n;



long long meme[100007][5];





long long solve(int index, int flag){

    if(index==n){

        return 0;

    }

    

    long long &res = meme[index][flag]; 

    

    if(res != -1){

        return res;

    }

    res = 0;

    

    if(flag!=0){

        res = max(res,vec[index][0]+solve(index+1,0));

    }

    if(flag!=1){

        res =  max(res,vec[index][1]+solve(index+1,1));

    }

    if(flag!=2){

        res =  max(res,vec[index][2]+solve(index+1,2));

    }

    return res;

}



int main(){

    cin>>n;

    int num1,num2,num3;

    memset(meme,-1,sizeof(meme));

    for(int i = 0 ; i<n ; i++){

        cin>>num1>>num2>>num3;

        vec[i].push_back(num1);

        vec[i].push_back(num2);

        vec[i].push_back(num3);

    }

    long long sol = solve(0,3);

    cout<<sol<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 7.953904420067528x):
#include <bits/stdc++.h>

 

#define f(i,a,b) for( ll i = a; i < b ; i++ )

#define af(i,a,b) for( ll i = a; i >= b ; i--)

#define rep(i,a,b,k) for(ll i = a; i < b ; i+= k )

#define arep(i,a,b,k) for( ll i = a; i >= b ; i-= k)

#define ones(x) (ll) __builtin_popcount(x)

#define fs first

#define sc second

#define pb push_back

#define po pop_back

#define mp make_pair  

#define sz(a) (ll) a.size()

#define all(a) a.begin(), a.end()

#define sor(a) sort( a.begin(), a.end() )

#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL)

#define ller ios::sync_with_stdio(false);cin.tsie(nullptr);cout.tie(nullptr)

#define watch(x) cout << (#x) << " is #" << (x) <<"#" <<endl

#define PI 3.1415926535

using namespace std;

typedef long long ll;

typedef long double ld; 

typedef pair<ll,ll> ii ;

typedef vector<ll>  vi ;

typedef vector<ii> vii ;

const ll mod = 1e9+7;

const ll MAX = 1e5+5;

const ll inf = 1e18+7;

ll dp[MAX][4];

ll a[MAX],b[MAX],c[MAX];

int main() {

  fastio;

  ll n;

  cin>>n;

  f(i,0,n) cin>>a[i]>>b[i]>>c[i];

  dp[0][0]=a[0];

  dp[0][1]=b[0];

  dp[0][2]=c[0];

  f(i,1,n){

    dp[i][0] = max(dp[i-1][1]+a[i],dp[i-1][2]+a[i]);

    dp[i][1] = max(dp[i-1][0]+b[i],dp[i-1][2]+b[i]);

    dp[i][2] = max(dp[i-1][0]+c[i],dp[i-1][1]+c[i]);

  }

  ll ans = max(dp[n-1][0],max(dp[n-1][1],dp[n-1][2]));

  cout<<ans<<"\n";

  return 0; 

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;



int ar[100010][5],n;

int dp[100010][5];



int call(int i,int f)

{



    if(i==n)

        return 0;



    if(dp[i][f]!=-1)

        return dp[i][f];



    int ans=0;



    if(f==0)

        ans+=max(ar[i][1]+call(i+1,1),ar[i][2]+call(i+1,2));



    else if(f==1)

        ans+=max(ar[i][0]+call(i+1,0),ar[i][2]+call(i+1,2));

    else

        ans+=max(ar[i][1]+call(i+1,1),ar[i][0]+call(i+1,0));



        dp[i][f]=ans;

        return ans;



}



int main()

{



    cin>>n;



    for(int i=0;i<n;i++)

        cin>>ar[i][0]>>ar[i][1]>>ar[i][2];



    memset(dp,-1,sizeof dp);



    int ans=max(max(call(0,0),call(0,1)),call(0,2));



    cout<<ans<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 2.9670403841655153x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

#define all(v) (v).begin(),(v).end()

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],b[N],c[N],n;

int solve(int p,int idx){

    if(idx==n)return 0;

    int &ret=dp[p][idx];

    if(ret)return ret;

    if(p==0)ret=a[idx];

    else if(p==1)ret=b[idx];

    else ret=c[idx];

    idx++;

    if(p==0)ret+=max(solve(1,idx),solve(2,idx));

    else if(p==1)ret+=max(solve(0,idx),solve(2,idx));

    else if(p==2)ret+=max(solve(0,idx),solve(1,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

    for(int i=0;i<n;i++){

        cin>>a[i]>>b[i]>>c[i];

    }

    int ans;

    ans=max({solve(0,0),solve(1,0),solve(2,0)});

    cout<<ans;

    return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;



#define MAX 100000



int dp[MAX+5][4];

int val[MAX+5][4];

int n;



void initialize()

{

    for(int i=0;i<=MAX;i++)

        for(int j=0;j<=3;j++)

            dp[i][j] = -1;

}



int vacation(int i, int j)

{

    if(i<=0) return 0;



  /*  if(i==1)

    {

        return val[i][j];

    }

*/

    if(dp[i][j]!=-1) return dp[i][j];



    int mx = INT_MIN;



    for(int k=1;k<=3;k++)

    {

        if(j==k) continue;



        mx = max(mx, vacation(i-1, k)+ val[i][j]);

    }





    dp[i][j] = mx;



    return mx;

}



int main()

{

    initialize();



    cin>>n;

    for(int i=1;i<=n;i++)

    {

        for(int j=1;j<=3;j++)

            cin>>val[i][j];

    }



    int retVal1 = vacation(n, 1);



    initialize();



    int retVal2 = vacation(n, 2);



    initialize();



    int retVal3 = vacation(n, 3);



    cout<<max({retVal1, retVal2, retVal3})<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.587007651573357x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

#define all(v) (v).begin(),(v).end()

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],b[N],c[N],n;

int solve(int p,int idx){

    if(idx==n)return 0;

    int &ret=dp[p][idx];

    if(ret)return ret;

    if(p==0)ret=a[idx];

    else if(p==1)ret=b[idx];

    else ret=c[idx];

    idx++;

    if(p==0)ret+=max(solve(1,idx),solve(2,idx));

    else if(p==1)ret+=max(solve(0,idx),solve(2,idx));

    else if(p==2)ret+=max(solve(0,idx),solve(1,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

    for(int i=0;i<n;i++){

        cin>>a[i]>>b[i]>>c[i];

    }

    int ans;

    ans=max({solve(0,0),solve(1,0),solve(2,0)});

    cout<<ans;

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int arr[100010][5];

int dp[100010][5];



int vac(int a, int b){

	if(a == 0) return 0;

	if(dp[a][b] != -1) return dp[a][b];

	

	for(int i = 1; i <= 3; i++){

		if(i != b){

			dp[a][b] = max(dp[a][b], vac(a - 1,i) + arr[a][i]);

		}

	}

	

	return dp[a][b];

}



int main (){

	memset(dp,-1,sizeof dp);

	int n;

	cin >> n;

	for(int i = 1; i <=n; i++){

		cin >> arr[i][1] >> arr[i][2] >> arr[i][3];

	}

	

	

	int ans = 0;

	for(int i = 1; i <= 3; i++){

		ans = max(ans , vac(n - 1, i) + arr[n][i]);

	}

	

	cout << ans << endl;

	

}

// Faster C++ solution for the same problem (measured speedup: 2.9698319980887464x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

#define all(v) (v).begin(),(v).end()

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],b[N],c[N],n;

int solve(int p,int idx){

    if(idx==n)return 0;

    int &ret=dp[p][idx];

    if(ret)return ret;

    if(p==0)ret=a[idx];

    else if(p==1)ret=b[idx];

    else ret=c[idx];

    idx++;

    if(p==0)ret+=max(solve(1,idx),solve(2,idx));

    else if(p==1)ret+=max(solve(0,idx),solve(2,idx));

    else if(p==2)ret+=max(solve(0,idx),solve(1,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

    for(int i=0;i<n;i++){

        cin>>a[i]>>b[i]>>c[i];

    }

    int ans;

    ans=max({solve(0,0),solve(1,0),solve(2,0)});

    cout<<ans;

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h> 

using namespace std;

#define INF INT_MAX

#define N 100005

#define K 3

int n, k;

int arr[N][K];

int dp[N][K];

int solve(int idx, int last_k){

    int res;

    if(idx == n)

        return 0;

    if(~last_k && ~dp[idx][last_k])

        return dp[idx][last_k];

    for(int i=0; i<K; ++i){

        if(last_k != i){

            dp[idx][last_k] = max(dp[idx][last_k], arr[idx][i]+solve(idx+1, i));

        }

    }

    return dp[idx][last_k];

}

int main()

{

    cin >> n;

    memset(dp, -1, sizeof(dp[0][0]) * N * K);

    for(int i=0; i<n; ++i)

        for(int j=0; j<3; ++j)

            cin >> arr[i][j];

    cout << solve(0, -1);

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 2.1695674964148384x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

#define all(v) (v).begin(),(v).end()

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],b[N],c[N],n;

int solve(int p,int idx){

    if(idx==n)return 0;

    int &ret=dp[p][idx];

    if(ret)return ret;

    if(p==0)ret=a[idx];

    else if(p==1)ret=b[idx];

    else ret=c[idx];

    idx++;

    if(p==0)ret+=max(solve(1,idx),solve(2,idx));

    else if(p==1)ret+=max(solve(0,idx),solve(2,idx));

    else if(p==2)ret+=max(solve(0,idx),solve(1,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

    for(int i=0;i<n;i++){

        cin>>a[i]>>b[i]>>c[i];

    }

    int ans;

    ans=max({solve(0,0),solve(1,0),solve(2,0)});

    cout<<ans;

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

typedef long long ll;

using namespace std;

const int N = 2e5+10 , M = 5e5+10, OO = 0x3f3f3f3f ,  MOD = 1e9+7;

const double EPS = 1e-3;

#define M_PI 3.14159265358979323846

#define clr(arr,val) memset(arr , val , sizeof (arr))

#define loop(i,n) for(int i = 0;i < int(n);i++)

#define rloop(i,n) for(int i = int(n)-1;i >= 0;i--)

#define range(i,a,b) for(int i = int(a);i <= int(b);i++)

#define pb push_back

#define F first

#define S second

#define all(v) ((v).begin()) , ((v).end())

#define sz(v)  ((int)((v).size()))

typedef pair <int,int> pii ;

typedef vector < pii > vii ;

typedef vector < int > vi;

int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction

//int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 direction

//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction

//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction

//freopen("input.txt", "r", stdin);

//freopen("output.txt", "w", stdout);







int n,x;

int a[N],b[N],c[N];

ll dp[N][4];





ll solve(int idx, int preC){ //preColour = (1 : a) -- (2 : b) -- (3 : c) -- (0 : invalid)



    if(idx == n) return 0;



    ll& ret = dp[idx][preC];

    if(~ret) return ret;



    ret = -OO;

    if(preC != 1) ret = max(ret, solve(idx+1 , 1)+a[idx]);

    if(preC != 2) ret = max(ret, solve(idx+1 , 2)+b[idx]);

    if(preC != 3) ret = max(ret, solve(idx+1 , 3)+c[idx]);



    return ret;



}







int main(){





    //ios_base::sync_with_stdio(0);

    //cin.tie(0);cout.tie(0);





    scanf("%d",&n);

    for(int i = 0 ; i < n ; ++i)

        scanf("%d %d %d",a+i,b+i,c+i);



    memset(dp, -1 , sizeof dp);

    ll ans = solve(0 , 0);

    printf("%lld\n",ans);







    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.546162762180811x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

#define all(v) (v).begin(),(v).end()

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],b[N],c[N],n;

int solve(int p,int idx){

    if(idx==n)return 0;

    int &ret=dp[p][idx];

    if(ret)return ret;

    if(p==0)ret=a[idx];

    else if(p==1)ret=b[idx];

    else ret=c[idx];

    idx++;

    if(p==0)ret+=max(solve(1,idx),solve(2,idx));

    else if(p==1)ret+=max(solve(0,idx),solve(2,idx));

    else if(p==2)ret+=max(solve(0,idx),solve(1,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

    for(int i=0;i<n;i++){

        cin>>a[i]>>b[i]>>c[i];

    }

    int ans;

    ans=max({solve(0,0),solve(1,0),solve(2,0)});

    cout<<ans;

    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cstring>

using namespace std;

const int maxn=1e5+5;

const int inf=0x3f3f3f3f;

int dp[maxn][4];

int a[maxn],b[maxn],c[maxn];

int main()

{

   int n;

   cin>>n;

   memset(dp,0,sizeof(dp));

   for(int i=1;i<=n;++i){

       cin>>a[i]>>b[i]>>c[i];

   }

   dp[1][1]=a[1];

   dp[1][2]=b[1];

   dp[1][3]=c[1];

   for(int i=2;i<=n;++i){

       dp[i][1]=max(dp[i-1][2]+a[i],dp[i-1][3]+a[i]);

       dp[i][2]=max(dp[i-1][1]+b[i],dp[i-1][3]+b[i]);

       dp[i][3]=max(dp[i-1][2]+c[i],dp[i-1][1]+c[i]);

   }

   cout<<max(dp[n][1],max(dp[n][2],dp[n][3]))<<endl;

   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.5739295005247578x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

#define all(v) (v).begin(),(v).end()

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],b[N],c[N],n;

int solve(int p,int idx){

    if(idx==n)return 0;

    int &ret=dp[p][idx];

    if(ret)return ret;

    if(p==0)ret=a[idx];

    else if(p==1)ret=b[idx];

    else ret=c[idx];

    idx++;

    if(p==0)ret+=max(solve(1,idx),solve(2,idx));

    else if(p==1)ret+=max(solve(0,idx),solve(2,idx));

    else if(p==2)ret+=max(solve(0,idx),solve(1,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

    for(int i=0;i<n;i++){

        cin>>a[i]>>b[i]>>c[i];

    }

    int ans;

    ans=max({solve(0,0),solve(1,0),solve(2,0)});

    cout<<ans;

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

//const int mod = 1e9+9;

/*ll gcd(ll x, ll y){return (y ? gcd(y, x%y) : x); }

*ll lcm(ll x, ll y){return x*(y/gcd(x,y));}

*void exEuclid(int a, int b){

	if(b==0){x = 1; y=0; d=a; return ;}

	exEuclid(b, a%b);

	int x1=y;

	int y1=x-(a/b)*y;

	x=x1; 100000

	y=y1;

} 

* */

//===========================================================//

int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i){

	if(i == n) return 0;

	if(dp[a][i] != -1) return dp[a][i];

	int& ans = dp[a][i];

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>n;

	for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

	cout<<max(go(0,0),max(go(1,0),go(2,0)))<<"\n";

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.180764070109373x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

#define all(v) (v).begin(),(v).end()

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],b[N],c[N],n;

int solve(int p,int idx){

    if(idx==n)return 0;

    int &ret=dp[p][idx];

    if(ret)return ret;

    if(p==0)ret=a[idx];

    else if(p==1)ret=b[idx];

    else ret=c[idx];

    idx++;

    if(p==0)ret+=max(solve(1,idx),solve(2,idx));

    else if(p==1)ret+=max(solve(0,idx),solve(2,idx));

    else if(p==2)ret+=max(solve(0,idx),solve(1,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

    for(int i=0;i<n;i++){

        cin>>a[i]>>b[i]>>c[i];

    }

    int ans;

    ans=max({solve(0,0),solve(1,0),solve(2,0)});

    cout<<ans;

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

const int oo = 1e9;



int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i)

{

	if(i==n) return 0;

	if(dp[a][i] != -1) return dp[a][i];

	int& ans = dp[a][i];

	if(a==0)

	{

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1)

	{

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2)

	{

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

	

}

int main()

{

    cin>>n;

    for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

    memset(dp, -1, sizeof dp);

    cout<<max(go(0,0), max(go(1,0),go(2,0)))<<"\n";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.1807420495764562x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

#define all(v) (v).begin(),(v).end()

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],b[N],c[N],n;

int solve(int p,int idx){

    if(idx==n)return 0;

    int &ret=dp[p][idx];

    if(ret)return ret;

    if(p==0)ret=a[idx];

    else if(p==1)ret=b[idx];

    else ret=c[idx];

    idx++;

    if(p==0)ret+=max(solve(1,idx),solve(2,idx));

    else if(p==1)ret+=max(solve(0,idx),solve(2,idx));

    else if(p==2)ret+=max(solve(0,idx),solve(1,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

    for(int i=0;i<n;i++){

        cin>>a[i]>>b[i]>>c[i];

    }

    int ans;

    ans=max({solve(0,0),solve(1,0),solve(2,0)});

    cout<<ans;

    return 0;

}