stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
/******

author :- Mahmoud Tarek.

*******/



#include<bits/stdc++.h>



using namespace std ;



#define rg register

#define dbg(v) cerr << #v << " " << v ;

typedef long long ll ;



int n;

struct doW{

int w[5] ;

};



vector<doW> a ;

int mem[100002][6] ;



int solve(int i, int ch){

if(i == n) return 0 ;

int & ret = mem[i][ch] ;

if(~ret) return ret ;

ret = 0 ;

for(int j = 2 ; j <= 4 ; j++){

    if(ch != j){

        ret = max(ret, a[i].w[j] + solve(i + 1, j)) ;

    }

}

return ret ;

}



int main() {

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

    cin >> n ;

    a.resize(n) ;

    memset(mem, -1, sizeof mem) ;

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

        for(int j = 2 ; j <= 4 ; j++) cin >> a[i].w[j] ;

    }

    cout << solve(0, 1) ;



}



/*

  Empty set of troubles.

*/


// Faster C++ solution for the same problem (measured speedup: 3.4032029152103904x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

using namespace std;



const int N=1e5+30;

typedef long long ll;

int n;

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

ll dp[N][5];

ll solve(int i,int last){

    if(i==n)return 0;

    int path1=-1 ,path2=-1,path3=-1;

    ll &ret=dp[i][last];

    if(~ret)return ret;

    if(last!=1){

        path1=a[i]+solve(i+1,1);

    }

    if(last!=2){

        path2=b[i]+solve(i+1,2);

    }

    if(last!=3){

        path3=c[i]+solve(i+1,3);

    }

    return ret=max(path1,max(path2,path3));



}

int main()

{

    memset(dp,-1,sizeof dp);

    cin>>n;

    for(int i=0;i<n;i++)cin>>a[i]>>b[i]>>c[i];

    cout<<solve(0,4);



}




// Faster C++ solution for the same problem (measured speedup: 5.0782458030289765x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







// */
// Slower C++ solution:
/* MaTb3aa */

#include<bits/stdc++.h>

#define ll long long

using namespace std;

int n;

const int N = 100005;

vector<int>v;

ll dp[100005][4];

int A[N], B[N], C[N];

ll solve(int i,int lst) {

	if (i == n)return 0;

	ll& ret = dp[i][lst];

	if (~ret)return ret;

	if (lst != 1)ret = max(ret, solve(i + 1, 1) + A[i]);

	if (lst != 2)ret = max(ret, solve(i + 1, 2) + B[i]);

	if (lst != 3)ret = max(ret, solve(i + 1, 3) + C[i]);

	return ret;

}

int main() {

	cin >> n;

	for (int i = 0; i < n; i++)cin >> A[i] >> B[i] >> C[i];

	memset(dp, -1, sizeof dp);

	cout << solve(0, 0);

	

		 

}

// Faster C++ solution for the same problem (measured speedup: 4.213254216076687x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

using namespace std;

int n;

int arr[100005][3];



int dp[100005][4];



int solve(int i, int prev){

	if(i == n) return 0;

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



	dp[i][prev] = -1e7;

	int a = 0, b = 0, c = 0;

	if(prev != 0)

		a = solve(i+1, 0) + arr[i][0];

	if(prev != 1)

		b = solve(i+1, 1) + arr[i][1];

	if(prev != 2)

		c = solve(i+1, 2) + arr[i][2];



	dp[i][prev] = max(a, max(b, c));

	return dp[i][prev];

}



int main(){

	cin>>n;

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

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



	memset(dp, -1, sizeof dp);

	cout<<solve(0, 3)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6043471742756896x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







// */
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

	cin.tie(NULL), cout.tie(NULL);

}

struct s {

	int x, y, z;

};



int n;

int v[(int)1e5][4];

int dp[(int)1e5][4];



int solve(int i = 0, int pre = 4) {

	if (i >= n)return 0;

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

	int ch1, ch2, ch3;

	ch1 = ch2 = ch3 = 0;

	if (pre != 1) {

		ch1 += solve(i + 1, 1) + v[i][1];

	}

	if (pre != 2) {

		ch2 += solve(i + 1, 2) + v[i][2];

	}

	if (pre != 3) {

		ch3 += solve(i + 1, 3) + v[i][3];

	}

	return dp[i][pre] = max({ ch1,ch2,ch3 });

}





int main()

{

	fast();



	cin >> n;

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

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

	}

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

	cout << solve();



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6160884884564375x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

#include<bits/stdc++.h>

using namespace std;



typedef long long ll;



int n;

int a[100005][4];

ll dp[100005][4];



ll solve(int d,int l){

    if(d==n+1) return 0;

    if(dp[d][l]!=-1) return dp[d][l];

    ll ret=0;

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

        if(i!=l){

            ret=max(ret, solve(d+1,i)+a[d][i]);

        }

    }

    return dp[d][l]=ret;



}





int main()

{

    cin>>n;

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

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

            cin>>a[i][j];

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

    cout<<solve(1,-1);



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.213346618313719x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

using namespace std;



typedef long double LD;

typedef long long ll;

typedef map<int,int> mii;

typedef pair<int,int> pii;

typedef vector<pii> vpii;



#define pb              push_back

#define mp              make_pair

#define F               first

#define S               second

#define mod             1000000007

#define INF             2147483647

#define Endl            "\n"

#define FasterIO        ios_base::sync_with_stdio(0);cin.tie(0); cout.tie(0) ;

#define precision(a)    cout << fixed << setprecision(a)

#define bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define reset(a,b)      memset(a,b,sizeof(a));

#define srt(a)      sort(a.begin(),a.end());



const int MAXN = 1000005;

const int SQRTN = 1003;

const int LOGN = 22;

const double PI=acos(-1);

const int sz=1e5+9;



int strtoint(string str){

    stringstream ss(str);

    int x = 0;

    ss >> x ;

    return x ;

}

string intostr(int x){

    stringstream ss;

    ss << x;

    string str = ss.str();

    return str;

}

void rem(string &s, const string &c) {

	auto pos = s.find(c);

	if (pos == string::npos) {

		cout << -1 << endl;

		exit(0);

	}

	s.erase(0, pos + 1);

}

void println (int n){

    cout<<n<<"\n";

}

void printvec(vector <int> &a){

    for (int i=0;i<a.size();i++){

        cout<<a[i]<<" ";

    }

    cout<<"\n";

}



void printgraph(vector <int> g[],int e){

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

        cout<<i<<" --> ";

        for (int j=0;j<g[i].size();j++){

            cout<<g[i][j]<<" ";

        }

        cout<<"\n";

    }

}



void printmap (map <int,int> &mymap){

    for (auto itr = mymap.begin(); itr != mymap.end(); ++itr) {

        cout << '\t' << itr->first

             << '\t' << itr->second << '\n';

    }

    cout << "\n";

}



int a[100005][4], dp[100005][4];

int N;



int rec(int pos, int type){



    if(pos>N) return 0;



    if(dp[pos][type]!=-1) return dp[pos][type];



    int ret = -1;

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

        if(i!=type)

            ret = max (ret, rec(pos+1,i) + a[pos+1][i]);



    return dp[pos][type] = ret;

}





void solve(){

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

    int i,j;

    cin>>N;

    for(i=1; i<=N; i++)

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

            cin>>a[i][j];





   cout<< rec(0,0) << "\n";



    //printvec(v);

}





int main(){

    //FasterIO;



    solve();

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6252666736151675x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

#define inf INFINITY

#define file freopen( "input.txt" , "r" , stdin) ; freopen( "output.txt" , "w" , stdout) ;

#define Gamed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define _3102_ main()

#define endl '\n'

typedef long long ll;

typedef unsigned long long ull;

typedef long double ld;

using namespace std;

const int N = 1e5 + 100;

const ll MX = 1e9;

const ll mod = 1e9 + 7;

const double eps = 1e-9 ;



ll n, a[N] , b[N] , c[N], dp[N][3] ;



ll solve( int i , int j ){

	if ( i < 0 ) return 0 ;

	ll &ret = dp[i][j] ;

	if ( ret != -1) return ret ;

	if ( j == 0 ){

		return ret = a[i] + max ( solve(i-1 , 1), solve(i-1,2) ) ;

	}

	else if ( j == 1 ){

		return ret = b[i] + max ( solve(i-1 , 0), solve(i-1,2) ) ;

	}

	return ret = c[i] + max ( solve(i-1 , 1), solve(i-1,0) ) ;

}



int _3102_{

	Gamed

	cin >> n ;

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

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

	}

	memset ( dp , -1 , sizeof dp);

	dp[0][0] = a[0] , dp[0][1] = b[0] , dp[0][2] = c[0] ;

	cout << max ( solve ( n - 1 , 1) , max ( solve ( n -1 , 0 ) , solve ( n - 1 , 2 ))) << endl ;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.4147791281931883x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

using namespace std;

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

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

typedef long long ll;

typedef unsigned long long ull;

#define FIN ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)

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

#define forn(i, n) forr(i, 0, n)

#define pb push_back

#define mp make_pair

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

#define DBG(x) cerr << #x << " = " << (x) << endl

#define MOD 1000000007

#define MAXN 100005

ll dp[MAXN][3];

void solve(vector< vector<ll> > &a, int n)

{

    forn(i,MAXN)

    {

        forn(j,3)

        {

            dp[i][j]=(pow(10,9)+5)*(ll)(-1);

        }

    }

    dp[0][0]=a[0][0], dp[0][1]=a[0][1], dp[0][2]=a[0][2];

    forr(i,1,n)

    {

        forn(j,3)

        {

            forn(k,3)

            {

                if(j==k)

                {

                    continue;

                }

                else dp[i][j]=max(dp[i][j],dp[i-1][k]+a[i][j]);

            }

        }

 

    }

    cout << max(dp[n-1][0],max(dp[n-1][1],dp[n-1][2])) << "\n";

}

 

int main() {

    FIN;

    int n;

    cin >> n;

    vector< vector<ll> > arreglo(n);

    forn(i,n) arreglo[i].resize(3);

    forn(i,n)

    {

        forn(j,3) cin >> arreglo[i][j];

    }

    solve(arreglo, n);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.4179429587064907x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

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

#define ll long long

using namespace std;

const int N = 100010;

int dp[N][3];

int main(){

    fast;

	int n, i, a, b, c;

	cin >> n;

	memset(dp, 0, sizeof dp);

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

		cin >> a >> b >> c;

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

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

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

	}

	cout << max(dp[n][0], max(dp[n][1], dp[n][2])) << "\n";

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.220761142193672x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

using namespace std;

#define tpl  tuple<int, int, int>

#define mt   make_tuple

#define LIM  100000





int n, dp[LIM+5][4], vis[LIM+5][4];

vector<tpl> v;



int f(int pos, int k){

    if(pos >= n) return 0;

    if(vis[pos][k]) return dp[pos][k];

    vis[pos][k] = 1;

    int ans;

    if(k == 0) ans = get<0>(v[pos]) + max(f(pos+1, 1), f(pos+1, 2));

    else if(k == 1) ans = get<1>(v[pos]) + max(f(pos+1, 0), f(pos+1, 2));

    else ans = get<2>(v[pos]) + max(f(pos+1, 0), f(pos+1, 1));



    return dp[pos][k] = ans;

}



int main(){

    scanf("%d", &n);

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

        int a, b, c; scanf("%d %d %d", &a, &b, &c);

        v.push_back(mt(a, b, c));

    }

    int ans = f(0, 0);

    memset(dp, 0, sizeof dp);

    memset(vis, 0, sizeof vis);

    ans = max(ans, f(0, 1));

    memset(dp, 0, sizeof dp);

    memset(vis, 0, sizeof vis);

    ans = max(ans, f(0, 2));



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

}


// Faster C++ solution for the same problem (measured speedup: 6.528905176697382x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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

#define ll long long

#define endl "\n"

#define shalaby ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);cerr.tie(0);

#define mod 100000000

using namespace std;



inline int D()

{

    int t;

    scanf("%d",&t);

    return t;

}

inline ll llD()

{

    ll t;

    scanf("%lld",&t);

    return t;

}

const int Max=1e5+8;

const int OO=-1e6;

int xt[4]= {1,0,0,-1};

int yt[4]= {0,1,-1,0};

////////////////////////////////////////////////////////////////////////////////////



ll mem[100005][5],a[100005],b[100005],c[100005],n,m;

bool valid(int x,int y)

{

    return x>-1 && y>-1 && x<n && y<m;

}

void init()

{

    memset(mem,-1,sizeof mem);

}

/////////////////////////////////////////////////////////////////////

ll solve(int i=0, int k=3)

{

    if(i==n)

        return 0;

    ll &ret=mem[i][k];

    if(~ret)return ret;

    ret =0;

    if(k!=0)

        ret=max(ret,solve(i+1,0)+a[i]);



    if(k!=1)

        ret=max(ret,solve(i+1,1)+b[i]);

    if(k!=2)

        ret=max(ret,solve(i+1,2)+c[i]);

    return ret;

}



int main()

{

    init();

    int T=1;

///   T=D();

    while(T-->0)

    {

        cin >> n ;

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

        {

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

        }

        cout << solve();

    }

}


// Faster C++ solution for the same problem (measured speedup: 5.078034146009375x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







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



#define pb push_back

#define mp make_pair



using namespace std;



typedef vector<int> vi;

typedef pair<int, int> ii;

typedef vector<ii> vii;

typedef unsigned long long ull;

typedef long long ll;





int DP[100005][3];

int V[100005][3];

bool ok[100005][3];

int N;



int dp(int i, int m) {

    if (ok[i][m]) {

        return DP[i][m];

    }

    

    if (i == N+1) {

        return 0;

    } else {

        int res = 0;

        if (m == 0) {

            res = max(dp(i+1, 1) + V[i][1], dp(i+1, 2) + V[i][2]);

        } else if (m == 1) {

            res = max(dp(i+1, 0) + V[i][0], dp(i+1, 2) + V[i][2]);

        } else {

            res = max(dp(i+1, 0) + V[i][0], dp(i+1, 1) + V[i][1]);

        }

        

        ok[i][m] = true;

        DP[i][m] = res;

        return DP[i][m];

    }

}



int main() {

    std::ios::sync_with_stdio(false);

    cin.tie(nullptr);

    cout.tie(nullptr);

    

    cin >> N;

    

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

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

    }

    

    int res = 0;

    res = dp(1, 1);

    memset(DP, 0, sizeof(DP));

    memset(ok, 0, sizeof(ok));

    res = max(res, dp(1, 2));

    memset(DP, 0, sizeof(DP));

    memset(ok, 0, sizeof(ok));

    res = max(res, dp(1, 3));

    

    cout << res << endl;

    

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.9953215732967156x):
 // /*

#include <bits/stdc++.h>

using namespace std;

int dp [100003][4];

int n;

int arr [100003];

int brr [100003];

int crr [100003];

int main ()

{



cin >>n;

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



{

cin >> arr [i]>>brr [i]>>crr [i];

}

dp [0][0] = arr [0];

dp [0][1] = brr [0];

dp [0][2] = crr [0];

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

{

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

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

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

}

cout << max ({dp [n - 1][0],dp [n - 1][1],dp [n - 1][2]})<<'\n';

}







// */