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

using namespace std ;

#define ll long long

inline int D(){

    int t ;

    scanf( "%d" , &t ) ;

    return t ;

}

inline ll LLD(){

    ll t ;

    scanf( "%lld" , &t ) ;

    return t ;

}

//################################################################

const int N=100009;

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

int mem[N][4];

int solve(int ind ,int lastInd){

    if(ind==n)return 0;

    int  ret=INT_MIN;

    if(mem[ind][lastInd]!=-1)return mem[ind][lastInd];

    if(lastInd!=0){



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

    }

    if(lastInd!=1){



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

    }

    if(lastInd!=2){



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

    }

    return mem[ind][lastInd]=ret;



}

int main() {

    n=D();

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

        a[i]=D(),b[i]=D(),c[i]=D();

    }

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

    printf("%d",solve(0,3));





}

// Faster C++ solution for the same problem (measured speedup: 2.5574889261272045x):
//C 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int raw[100000][10],dp[100000][10];

int main(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	int n,m,k;

	cin >> n;

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

		for(int j = 1;j <= 3;j++)cin >> raw[i][j];

	}

	ll ans = 0;

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

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

			for(int k = 1;k <= 3;k++)if(k != j){

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

			}

			dp[i][j] += raw[i][j];

		}

	}

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

	return 0;

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

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int N = 1e5+10 , OO = 0x3f3f3f3f ;

//const int mod = 1000000007;

 

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

double startTime;

double getCurrentTime() {

	return ((double)clock() - startTime) / CLOCKS_PER_SEC;

}

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

#define range(vec) for(auto &x : vec) cin >> x;

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

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

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

int dp[N][4];



int solve(int idx , int lst){

	

	if(idx == n) return 0;

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

	if(~ret) return ret;

	ret = -OO;

	if(lst != 0) // swim

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

	if(lst != 1)

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

	if(lst != 2)

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

		

	return ret;

}





int main()

{

    startTime = (double)clock();

    #ifndef ONLINE_JUDGE

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

        //freopen("out.in", "w", stdout);

    #endif

	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

	

	cin >> n;

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

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



	clr(dp , -1);

	cout << solve(0 , 3) << "\n";





    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.5973990989007247x):
//C 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int raw[100000][10],dp[100000][10];

int main(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	int n,m,k;

	cin >> n;

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

		for(int j = 1;j <= 3;j++)cin >> raw[i][j];

	}

	ll ans = 0;

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

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

			for(int k = 1;k <= 3;k++)if(k != j){

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

			}

			dp[i][j] += raw[i][j];

		}

	}

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

	return 0;

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

#define F first

#define S second

#define T second.first

#define C second.second

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

using namespace std;

typedef long long ll;

typedef pair<int, int> pii;

typedef pair<int,pair<int,int>> piii;



const int N= 1e5+5, L = 14, OO = 0x3f3f3f3f;

const ll M=10 ;

const double ESP = 1e-8;



int n, k;

piii A[N];

int dp[N][5];

int solve(int idx, int prev){

    if(idx == n)

        return 0;



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

    if(~ret)

        return ret;



    ret = max(ret, (prev != 1)? solve(idx+1, 1)+A[idx].F : -1);

    ret = max(ret, (prev != 2)? solve(idx+1, 2)+A[idx].T : -1);

    return ret = max(ret, (prev != 3)? solve(idx+1, 3)+A[idx].C : -1);



}



int main(){

    FIO

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

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

    cin>>n;

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

    memset(dp, -1, sizeof dp);

    cout<<solve(0,4);

}


// Faster C++ solution for the same problem (measured speedup: 2.9806193220643995x):
//C 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int raw[100000][10],dp[100000][10];

int main(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	int n,m,k;

	cin >> n;

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

		for(int j = 1;j <= 3;j++)cin >> raw[i][j];

	}

	ll ans = 0;

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

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

			for(int k = 1;k <= 3;k++)if(k != j){

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

			}

			dp[i][j] += raw[i][j];

		}

	}

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

	return 0;

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



using namespace std;



const int N=1e5;

int n;

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

int mem[N][5];



int solve(int i,int lst){

    if(i==n)

        return 0;

    if(mem[i][lst]!=-1) return mem[i][lst];

    int op1=lst!=0?solve(i+1,0)+a[i]:-1000000000;

    int op2=lst!=1?solve(i+1,1)+b[i]:-1000000000;

    int op3=lst!=2?solve(i+1,2)+c[i]:-1000000000;

    return mem[i][lst]=max(op1,max(op2,op3));

}



int main() {

    scanf("%d",&n);

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

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

    memset(mem,-1,sizeof mem);

    printf("%d",max(solve(0,0),max(solve(0,1),solve(0,2))));

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.958700324204092x):
//C 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int raw[100000][10],dp[100000][10];

int main(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	int n,m,k;

	cin >> n;

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

		for(int j = 1;j <= 3;j++)cin >> raw[i][j];

	}

	ll ans = 0;

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

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

			for(int k = 1;k <= 3;k++)if(k != j){

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

			}

			dp[i][j] += raw[i][j];

		}

	}

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

	return 0;

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

using namespace std;

#define ll long long

#define ull unsigned long long

#define PII pair<int,int>

#define PSI pair<string,int>

#define PIII pair< int, pair<int,int> >

const int MXN = 1e5+5;

int a[MXN][4], dp[MXN][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 read(){



    int i,j;

    cin>>N;

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

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

            cin>>a[i][j];

}



void solve(){



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

    read();

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

}



int main(){



    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);

    solve();

}


// Faster C++ solution for the same problem (measured speedup: 2.585710615980025x):
//C 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int raw[100000][10],dp[100000][10];

int main(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	int n,m,k;

	cin >> n;

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

		for(int j = 1;j <= 3;j++)cin >> raw[i][j];

	}

	ll ans = 0;

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

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

			for(int k = 1;k <= 3;k++)if(k != j){

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

			}

			dp[i][j] += raw[i][j];

		}

	}

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

	return 0;

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

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

int m[MXN][3];

int dp[MXN][4];



int x[3] = {0, 1, 2};



int Dp(int d, int lst) {

    if(d == n) return 0;

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

    int ans = 0;

    for(int i = 0; i < 3; i++) {

        if(lst != x[i]) {

            ans = max(ans, Dp(d + 1, i) + m[d][i]);

        }

    }

    return dp[d][lst] = ans;

}



int main() {

    ios_base::sync_with_stdio(false); cin.tie(NULL);

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.586548098783764x):
//C 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int raw[100000][10],dp[100000][10];

int main(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	int n,m,k;

	cin >> n;

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

		for(int j = 1;j <= 3;j++)cin >> raw[i][j];

	}

	ll ans = 0;

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

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

			for(int k = 1;k <= 3;k++)if(k != j){

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

			}

			dp[i][j] += raw[i][j];

		}

	}

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

	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: 7.962896732264393x):
//C 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int raw[100000][10],dp[100000][10];

int main(){

	ios::sync_with_stdio(0);

	cin.tie(0);

	int n,m,k;

	cin >> n;

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

		for(int j = 1;j <= 3;j++)cin >> raw[i][j];

	}

	ll ans = 0;

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

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

			for(int k = 1;k <= 3;k++)if(k != j){

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

			}

			dp[i][j] += raw[i][j];

		}

	}

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

	return 0;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

int N,A[MAX],B[MAX],C[MAX],dp[MAX][3];

int solve(int idx,int pr)

{

    if(idx>N)return 0;

    if(dp[idx][pr]!=-1)return dp[idx][pr];

    int res=0;

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

        if(i==pr)continue;

        int val=0;

        if(i==1)val=A[idx];

        if(i==2)val=B[idx];

        if(i==3)val=C[idx];

        res = max(res,solve(idx+1,i)+val);

    }

    return dp[idx][pr]=res;

}

int  main ()

{

    memset(dp,-1,sizeof dp);

    cin >> N;

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

        cin >> A[i] >> B[i] >> C[i];

    }

    cout<<solve(1,0)<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 2.207162598500947x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int felicidad[100005][3];

ll memo[100005][3];

int N;



ll maxFeliz(int dia, int act){

	if(!dia) return felicidad[0][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = felicidad[dia][act]+

					max(maxFeliz(dia-1, (act+1)%3),

						maxFeliz(dia-1, (act+2)%3));

}



int feli[3];

ll mem [2][3];



ll calcFelicidad(){

	cin>>feli[0]>>feli[1]>>feli[2];

	mem[0][0] = feli[0];

	mem[0][1] = feli[1];

	mem[0][2] = feli[2];

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

		cin>>feli[0]>>feli[1]>>feli[2];

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

			mem[i&1][k] = feli[k]+max(mem[(i-1)&1][(k+1)%3],

								mem[(i-1)&1][(k+2)%3]);

	}

	return max(mem[(N-1)&1][0],

			   max(mem[(N-1)&1][1],mem[(N-1)&1][2]));

}





int main() {

	cin>>N;

	/*for(int i = 0; i<N; i++)

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

	cout<<max(maxFeliz(N-1,0), max(maxFeliz(N-1,1), maxFeliz(N-1,2)))<<"\n";*/

	cout<<calcFelicidad()<<"\n";

	return 0;

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

using namespace std;

#define print printf("==================\n")

#define ll long long

#define pi acos(-1.0)

#define eps 1e-16

#define max3(a,b,c) max(a,max(b,c))

const ll INF = 1 << 30;

typedef pair<ll, ll  > payar;

typedef struct

{

    ll  x, y;

} point;

priority_queue<payar, vector<payar>,  greater<payar> > pq; ///accending

vector< pair<ll,payar  > > vpp;

vector < payar >  vp;



ll n ;

int ara[100005][4];

ll dp[100005][4];

ll cal(int day, int itm)

{

    if(day>n)

    {

        return 0;

    }



    if(dp[day][itm]!=-1)

        return dp[day][itm];

    ll maxi = -1;

    if(day==1)

    {

   return   maxi = dp[day][itm]=max3( cal(day+1,1)+ara[day][1], cal(day+1,2)+ara[day][2], cal(day+1,3)+ara[day][3] );



    }

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

    {

            if(i==itm)

                continue;

            maxi=max( cal(day+1,i)+ara[day][i],maxi);





    }

    return dp[day][itm]=maxi;

}



int main()

{



    cin>>n;

    memset(dp,-1,sizeof dp);

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

    {

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

        {

            cin>>ara[i][j];

        }

    }

    cout<<cal(1,1);







    return  0 ;

}


// Faster C++ solution for the same problem (measured speedup: 4.215046674381924x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int felicidad[100005][3];

ll memo[100005][3];

int N;



ll maxFeliz(int dia, int act){

	if(!dia) return felicidad[0][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = felicidad[dia][act]+

					max(maxFeliz(dia-1, (act+1)%3),

						maxFeliz(dia-1, (act+2)%3));

}



int feli[3];

ll mem [2][3];



ll calcFelicidad(){

	cin>>feli[0]>>feli[1]>>feli[2];

	mem[0][0] = feli[0];

	mem[0][1] = feli[1];

	mem[0][2] = feli[2];

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

		cin>>feli[0]>>feli[1]>>feli[2];

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

			mem[i&1][k] = feli[k]+max(mem[(i-1)&1][(k+1)%3],

								mem[(i-1)&1][(k+2)%3]);

	}

	return max(mem[(N-1)&1][0],

			   max(mem[(N-1)&1][1],mem[(N-1)&1][2]));

}





int main() {

	cin>>N;

	/*for(int i = 0; i<N; i++)

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

	cout<<max(maxFeliz(N-1,0), max(maxFeliz(N-1,1), maxFeliz(N-1,2)))<<"\n";*/

	cout<<calcFelicidad()<<"\n";

	return 0;

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

#include <cstdint>

#include <cinttypes>

#include <cstring>

#include <algorithm>



using i32 = std::int_fast32_t;



const int MaxN = 100001;



int A[MaxN][3], dp[MaxN][3];



signed main(signed argc, char **argv) {

	i32 n;

	std::scanf("%" SCNdFAST32, &n);

	

	for (i32 i = 0; i < n; ++ i) {

		std::scanf("%" SCNdFAST32 "%" SCNdFAST32 "%" SCNdFAST32, A[i], A[i] + 1, A[i] + 2);

	}

	

	memset(dp, 0, sizeof dp);

	

	dp[0][0] = A[0][0];

	dp[0][1] = A[0][1];

	dp[0][2] = A[0][2];

	

	for (i32 i = 1; i < n; ++ i) {

		for (i32 j = 0; j < 3; ++ j) {

			for (i32 a = 0; a < 3; ++ a) {

				if (a == j) {

					continue;

				}

				dp[i][j] = std::max(dp[i][j], dp[i - 1][a] + A[i][j]);

			}

		}

	}

	

	std::printf("%" PRIdFAST32, std::max(dp[n - 1][0], std::max(dp[n - 1][1], dp[n - 1][2])));

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3873618585982075x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int felicidad[100005][3];

ll memo[100005][3];

int N;



ll maxFeliz(int dia, int act){

	if(!dia) return felicidad[0][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = felicidad[dia][act]+

					max(maxFeliz(dia-1, (act+1)%3),

						maxFeliz(dia-1, (act+2)%3));

}



int feli[3];

ll mem [2][3];



ll calcFelicidad(){

	cin>>feli[0]>>feli[1]>>feli[2];

	mem[0][0] = feli[0];

	mem[0][1] = feli[1];

	mem[0][2] = feli[2];

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

		cin>>feli[0]>>feli[1]>>feli[2];

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

			mem[i&1][k] = feli[k]+max(mem[(i-1)&1][(k+1)%3],

								mem[(i-1)&1][(k+2)%3]);

	}

	return max(mem[(N-1)&1][0],

			   max(mem[(N-1)&1][1],mem[(N-1)&1][2]));

}





int main() {

	cin>>N;

	/*for(int i = 0; i<N; i++)

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

	cout<<max(maxFeliz(N-1,0), max(maxFeliz(N-1,1), maxFeliz(N-1,2)))<<"\n";*/

	cout<<calcFelicidad()<<"\n";

	return 0;

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

using namespace std;

#define N (int)2e5

int n,a[N],b[N],c[N],memo[N][10];



// a = 0, b = 1, c = 2

int dp(int curr , int last) {

    if (curr > n) return 0;

    int &sol = memo[curr][last];

    if (sol != -1) return sol;

    if (last == 0) {

        sol = max(

            dp(curr + 1 , 1) + b[curr] ,

            dp(curr + 1 , 2) + c[curr]

        );

    } else if (last == 1) {

        sol = max(

            dp(curr + 1 , 0) + a[curr] ,

            dp(curr + 1 , 2) + c[curr]

        );

    } else {

        sol = max(

            dp(curr + 1 , 0) + a[curr] ,

            dp(curr + 1 , 1) + b[curr]

        );

    }

    return sol;

}



int main() {

    scanf("%d" , &n);

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

        scanf("%d%d%d" , &a[i] , &b[i] , &c[i]);

    memset (memo , -1 , sizeof memo);

    printf("%d\n" , max(dp(2 , 0) + a[1] , max(dp(2 , 1) + b[1] , dp(2 , 2) + c[1])));

}


// Faster C++ solution for the same problem (measured speedup: 9.350998212955306x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int felicidad[100005][3];

ll memo[100005][3];

int N;



ll maxFeliz(int dia, int act){

	if(!dia) return felicidad[0][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = felicidad[dia][act]+

					max(maxFeliz(dia-1, (act+1)%3),

						maxFeliz(dia-1, (act+2)%3));

}



int feli[3];

ll mem [2][3];



ll calcFelicidad(){

	cin>>feli[0]>>feli[1]>>feli[2];

	mem[0][0] = feli[0];

	mem[0][1] = feli[1];

	mem[0][2] = feli[2];

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

		cin>>feli[0]>>feli[1]>>feli[2];

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

			mem[i&1][k] = feli[k]+max(mem[(i-1)&1][(k+1)%3],

								mem[(i-1)&1][(k+2)%3]);

	}

	return max(mem[(N-1)&1][0],

			   max(mem[(N-1)&1][1],mem[(N-1)&1][2]));

}





int main() {

	cin>>N;

	/*for(int i = 0; i<N; i++)

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

	cout<<max(maxFeliz(N-1,0), max(maxFeliz(N-1,1), maxFeliz(N-1,2)))<<"\n";*/

	cout<<calcFelicidad()<<"\n";

	return 0;

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

using namespace std;

typedef long long LL;



int a[100005],b[100005],c[100005];

int dp[100007][5];



int Solve(int idx, int n, int prev){

	

	if(idx == n)return 0;

	

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

	

	int ans = -1;

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

		if(i == prev)continue;

		if(i == 1)ans = max(ans, a[idx] + Solve(idx+1, n, i));

		else if(i == 2)ans = max(ans, b[idx] + Solve(idx+1, n, i));

		else ans = max(ans, c[idx] + Solve(idx+1, n, i));		

	}

	return dp[idx][prev] = ans;

}



int main(){

	int n; cin>>n;

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

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

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

	}

	int ret = -1;

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

		ret = max(ret, Solve(0,n,i));

	}

	cout<<ret<<endl;



//	cout<<Solve(0,n,0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.002160004426059x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int felicidad[100005][3];

ll memo[100005][3];

int N;



ll maxFeliz(int dia, int act){

	if(!dia) return felicidad[0][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = felicidad[dia][act]+

					max(maxFeliz(dia-1, (act+1)%3),

						maxFeliz(dia-1, (act+2)%3));

}



int feli[3];

ll mem [2][3];



ll calcFelicidad(){

	cin>>feli[0]>>feli[1]>>feli[2];

	mem[0][0] = feli[0];

	mem[0][1] = feli[1];

	mem[0][2] = feli[2];

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

		cin>>feli[0]>>feli[1]>>feli[2];

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

			mem[i&1][k] = feli[k]+max(mem[(i-1)&1][(k+1)%3],

								mem[(i-1)&1][(k+2)%3]);

	}

	return max(mem[(N-1)&1][0],

			   max(mem[(N-1)&1][1],mem[(N-1)&1][2]));

}





int main() {

	cin>>N;

	/*for(int i = 0; i<N; i++)

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

	cout<<max(maxFeliz(N-1,0), max(maxFeliz(N-1,1), maxFeliz(N-1,2)))<<"\n";*/

	cout<<calcFelicidad()<<"\n";

	return 0;

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

#include <stdlib.h>

#include <string.h>

#include <math.h>

#include <queue>

#include <stack>

#include <vector>

#include <algorithm>

#include <iostream>



using namespace std;



const int MAXN = 1e5 + 5;

int n,m;

int a,b,c;

int d,e,f;

vector <int> A;

vector <int> B;

vector <int> C;

vector <int> dp[MAXN];



int main()

{

	scanf("%d",&n);

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

	{

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

		A.push_back(a);//第一天的a现在是a[0]

		B.push_back(b);

		C.push_back(c);

	}

	dp[0].push_back(A[0]);

	dp[0].push_back(B[0]);

	dp[0].push_back(C[0]);

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

	{

		d = max(dp[i - 1][1],dp[i - 1][2]);

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

		f = max(dp[i - 1][0],dp[i - 1][1]);

		dp[i].push_back(d + A[i]);

		dp[i].push_back(e + B[i]);

		dp[i].push_back(f + C[i]);

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

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

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

	}

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

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.515652122205404x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int felicidad[100005][3];

ll memo[100005][3];

int N;



ll maxFeliz(int dia, int act){

	if(!dia) return felicidad[0][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = felicidad[dia][act]+

					max(maxFeliz(dia-1, (act+1)%3),

						maxFeliz(dia-1, (act+2)%3));

}



int feli[3];

ll mem [2][3];



ll calcFelicidad(){

	cin>>feli[0]>>feli[1]>>feli[2];

	mem[0][0] = feli[0];

	mem[0][1] = feli[1];

	mem[0][2] = feli[2];

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

		cin>>feli[0]>>feli[1]>>feli[2];

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

			mem[i&1][k] = feli[k]+max(mem[(i-1)&1][(k+1)%3],

								mem[(i-1)&1][(k+2)%3]);

	}

	return max(mem[(N-1)&1][0],

			   max(mem[(N-1)&1][1],mem[(N-1)&1][2]));

}





int main() {

	cin>>N;

	/*for(int i = 0; i<N; i++)

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

	cout<<max(maxFeliz(N-1,0), max(maxFeliz(N-1,1), maxFeliz(N-1,2)))<<"\n";*/

	cout<<calcFelicidad()<<"\n";

	return 0;

}