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

using namespace std;



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



int n;

int arr[3][N];



int mem[N][4];

int solve(int i, int last)

{

    if(i==n)

        return 0;

    if(mem[i][last]!=-1)

        return mem[i][last];

    int op1= (last!=1) ? solve(i+1, 1)+arr[0][i] : 0;

    int op2= (last!=2) ? solve(i+1, 2)+arr[1][i] : 0;

    int op3= (last!=3) ? solve(i+1, 3)+arr[2][i] : 0;

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

}





int main()

{

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

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

    scanf("%d",&n);

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

        scanf("%d %d %d",arr[0]+i,arr[1]+i,arr[2]+i);

    memset(mem, -1, sizeof mem);

    printf("%d\n",solve(0,0));







    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.5635214740814467x):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#include <complex>

#define ll long long

#define  point complex<double>

#define  pi acos(-1)

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

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const ll N = 2e5 + 5, mod = 1e9 + 7, inff = -1e12,inf=1e12;

const double ep = 0.0000000001;

//const double pi = 3.14159265359;



double dot(point a, point b) {

	return  (conj(a)*b).real();

}

double cross(point a, point b) {

	return  (conj(a)*b).imag();

}

double angle(point a, point b)

{

	double dt = dot(a, b);

	dt /= abs(a);

	dt /= abs(b);

	return acos(dt) * 180.0 / pi;

}



bool isrightangel(int ind) {//90 degree

	vector<point>v;

	return(angle(v[ind] - v[(ind + 1) % 3], v[ind] - v[(ind + 2) % 3]) == 90);//bshof hal eldot producty3ny fe

	//dl3en 3moden 3ashn yb2a right trinagle

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

	if (exp == 0)

		return 1;

	ll ans = fp(base, exp / 2);

	ans = (ans * ans) % mod;

	if (exp % 2 != 0)

		ans = (ans * (base % mod)) % mod;

	return ans;

}



void calcFacAndInv(ll n)

{

	fact[0] = inv[0] = 1;

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

	{

		fact[i] = (i * fact[i - 1]) % mod;

		inv[i] = fp(fact[i], mod - 2);

	}

}



ll ncr(ll n, ll r)

{

	return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

	return (fact[n] * inv[n - r]) % mod;

}













	int dpa[100005], dpb[100005], dpc[100005];

int main()

{

	ll n,arr[100005],a,b,c;

	char tmp;

	cin >> n;

	

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

		cin >> dpa[i] >> dpb[i] >> dpc[i];

	}

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

		dpa[i] += max(dpb[i - 1], dpc[i - 1]);

		dpb[i] += max(dpa[i-1], dpc[i-1]);

		dpc[i] += max(dpb[i-1], dpa[i-1]);

	}

	cout << max(dpa[n - 1], max(dpb[n - 1], dpc[n - 1]));



	return 0;

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

using namespace std;

#define ALL(x) begin(x),end(x)

typedef long long ll;

int OO = 0x3f3f3f3f;



int n;

int a[100005];

int b[100005];

int c[100005];

int mem[100005][3];



int solve(int x, int k)

{

    int op1, op2, op3;

    if(x == n)

        return 0;



    if(mem[x][k] != -1)

        return mem[x][k];



    op1 = op2 = op3 = 0;

    if(k != 1)

        op1 = solve(x+1, 1)+a[x];

    if(k != 2)

        op2 = solve(x+1, 2)+b[x];

    if(k != 3)

        op3 = solve(x+1, 3)+c[x];



    return mem[x][k] = max(op1, max(op2, op3));

}



int main()

{

    memset(mem, -1, sizeof mem);

    cin >> n;

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

    {

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

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.189290457172422x):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#include <complex>

#define ll long long

#define  point complex<double>

#define  pi acos(-1)

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

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const ll N = 2e5 + 5, mod = 1e9 + 7, inff = -1e12,inf=1e12;

const double ep = 0.0000000001;

//const double pi = 3.14159265359;



double dot(point a, point b) {

	return  (conj(a)*b).real();

}

double cross(point a, point b) {

	return  (conj(a)*b).imag();

}

double angle(point a, point b)

{

	double dt = dot(a, b);

	dt /= abs(a);

	dt /= abs(b);

	return acos(dt) * 180.0 / pi;

}



bool isrightangel(int ind) {//90 degree

	vector<point>v;

	return(angle(v[ind] - v[(ind + 1) % 3], v[ind] - v[(ind + 2) % 3]) == 90);//bshof hal eldot producty3ny fe

	//dl3en 3moden 3ashn yb2a right trinagle

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

	if (exp == 0)

		return 1;

	ll ans = fp(base, exp / 2);

	ans = (ans * ans) % mod;

	if (exp % 2 != 0)

		ans = (ans * (base % mod)) % mod;

	return ans;

}



void calcFacAndInv(ll n)

{

	fact[0] = inv[0] = 1;

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

	{

		fact[i] = (i * fact[i - 1]) % mod;

		inv[i] = fp(fact[i], mod - 2);

	}

}



ll ncr(ll n, ll r)

{

	return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

	return (fact[n] * inv[n - r]) % mod;

}













	int dpa[100005], dpb[100005], dpc[100005];

int main()

{

	ll n,arr[100005],a,b,c;

	char tmp;

	cin >> n;

	

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

		cin >> dpa[i] >> dpb[i] >> dpc[i];

	}

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

		dpa[i] += max(dpb[i - 1], dpc[i - 1]);

		dpb[i] += max(dpa[i-1], dpc[i-1]);

		dpc[i] += max(dpb[i-1], dpa[i-1]);

	}

	cout << max(dpa[n - 1], max(dpb[n - 1], dpc[n - 1]));



	return 0;

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



using namespace std;



const int N = 1e5 + 9, OO = 0x3f3f3f3f;



int n, points[N][3], dp[N][4];



int solve(int i, int prev){

    if(i == n)  return 0;



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

    if(~ret)    return ret;



    ret = 0;

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

        if(prev != j)   ret = max(ret, points[i][j] + solve(i + 1, j));





    return ret;

}



int main(){



    cin >> n;

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

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



    memset(dp, -1, sizeof dp);



    cout << solve(0, 3);

}


// Faster C++ solution for the same problem (measured speedup: 2.578564939255718x):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#include <complex>

#define ll long long

#define  point complex<double>

#define  pi acos(-1)

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

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const ll N = 2e5 + 5, mod = 1e9 + 7, inff = -1e12,inf=1e12;

const double ep = 0.0000000001;

//const double pi = 3.14159265359;



double dot(point a, point b) {

	return  (conj(a)*b).real();

}

double cross(point a, point b) {

	return  (conj(a)*b).imag();

}

double angle(point a, point b)

{

	double dt = dot(a, b);

	dt /= abs(a);

	dt /= abs(b);

	return acos(dt) * 180.0 / pi;

}



bool isrightangel(int ind) {//90 degree

	vector<point>v;

	return(angle(v[ind] - v[(ind + 1) % 3], v[ind] - v[(ind + 2) % 3]) == 90);//bshof hal eldot producty3ny fe

	//dl3en 3moden 3ashn yb2a right trinagle

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

	if (exp == 0)

		return 1;

	ll ans = fp(base, exp / 2);

	ans = (ans * ans) % mod;

	if (exp % 2 != 0)

		ans = (ans * (base % mod)) % mod;

	return ans;

}



void calcFacAndInv(ll n)

{

	fact[0] = inv[0] = 1;

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

	{

		fact[i] = (i * fact[i - 1]) % mod;

		inv[i] = fp(fact[i], mod - 2);

	}

}



ll ncr(ll n, ll r)

{

	return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

	return (fact[n] * inv[n - r]) % mod;

}













	int dpa[100005], dpb[100005], dpc[100005];

int main()

{

	ll n,arr[100005],a,b,c;

	char tmp;

	cin >> n;

	

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

		cin >> dpa[i] >> dpb[i] >> dpc[i];

	}

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

		dpa[i] += max(dpb[i - 1], dpc[i - 1]);

		dpb[i] += max(dpa[i-1], dpc[i-1]);

		dpc[i] += max(dpb[i-1], dpa[i-1]);

	}

	cout << max(dpa[n - 1], max(dpb[n - 1], dpc[n - 1]));



	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: 1.3613436516291235x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

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

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

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

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

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

#define     Endl            "\n";

#define     tab             "\t"

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

#define     sf(a)           scanf(" %d", &a);

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

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz][3],dp[sz][3];



int clac(int i,int ignore ,int value){

    if(i>n)return value;

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

    int ret=0;

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

        if(j==ignore)

            continue;

        ret=max(ret,value+clac(i+1,j,arr[i][j]));

    }

    return dp[i][ignore]=ret;

}



void solve(){

    cin>>n;

    reset(dp,-1);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    lp(i,1,n+1)cin>>arr[i][0]>>arr[i][1]>>arr[i][2];

    int ans1=clac(2,0,arr[1][0]);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    int ans2=clac(2,1,arr[1][1]);

    int ans3=clac(2,2,arr[1][2]);

    //cout<<ans1<<tab<<ans2<<tab<<ans3<<endl;

    //int ans=max(ans1,(max(ans2,ans3)));

   // bug(ans1,1);

    cout<<max(ans1,(max(ans2,ans3)))<<"\n";

}



int main(){

    //FasterIO;

    #ifndef ONLINE_JUDGE

       //freopen("1.txt","r",stdin);

    #endif

    // int t;

    //cin>>t;

    //while(t--)

        solve();

}
// 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: 1.6457983251171264x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

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

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

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

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

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

#define     Endl            "\n";

#define     tab             "\t"

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

#define     sf(a)           scanf(" %d", &a);

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

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz][3],dp[sz][3];



int clac(int i,int ignore ,int value){

    if(i>n)return value;

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

    int ret=0;

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

        if(j==ignore)

            continue;

        ret=max(ret,value+clac(i+1,j,arr[i][j]));

    }

    return dp[i][ignore]=ret;

}



void solve(){

    cin>>n;

    reset(dp,-1);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    lp(i,1,n+1)cin>>arr[i][0]>>arr[i][1]>>arr[i][2];

    int ans1=clac(2,0,arr[1][0]);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    int ans2=clac(2,1,arr[1][1]);

    int ans3=clac(2,2,arr[1][2]);

    //cout<<ans1<<tab<<ans2<<tab<<ans3<<endl;

    //int ans=max(ans1,(max(ans2,ans3)));

   // bug(ans1,1);

    cout<<max(ans1,(max(ans2,ans3)))<<"\n";

}



int main(){

    //FasterIO;

    #ifndef ONLINE_JUDGE

       //freopen("1.txt","r",stdin);

    #endif

    // int t;

    //cin>>t;

    //while(t--)

        solve();

}
// 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: 1.362624505746402x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

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

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

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

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

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

#define     Endl            "\n";

#define     tab             "\t"

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

#define     sf(a)           scanf(" %d", &a);

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

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz][3],dp[sz][3];



int clac(int i,int ignore ,int value){

    if(i>n)return value;

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

    int ret=0;

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

        if(j==ignore)

            continue;

        ret=max(ret,value+clac(i+1,j,arr[i][j]));

    }

    return dp[i][ignore]=ret;

}



void solve(){

    cin>>n;

    reset(dp,-1);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    lp(i,1,n+1)cin>>arr[i][0]>>arr[i][1]>>arr[i][2];

    int ans1=clac(2,0,arr[1][0]);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    int ans2=clac(2,1,arr[1][1]);

    int ans3=clac(2,2,arr[1][2]);

    //cout<<ans1<<tab<<ans2<<tab<<ans3<<endl;

    //int ans=max(ans1,(max(ans2,ans3)));

   // bug(ans1,1);

    cout<<max(ans1,(max(ans2,ans3)))<<"\n";

}



int main(){

    //FasterIO;

    #ifndef ONLINE_JUDGE

       //freopen("1.txt","r",stdin);

    #endif

    // int t;

    //cin>>t;

    //while(t--)

        solve();

}
// 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: 1.1809756968528102x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

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

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

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

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

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

#define     Endl            "\n";

#define     tab             "\t"

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

#define     sf(a)           scanf(" %d", &a);

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

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz][3],dp[sz][3];



int clac(int i,int ignore ,int value){

    if(i>n)return value;

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

    int ret=0;

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

        if(j==ignore)

            continue;

        ret=max(ret,value+clac(i+1,j,arr[i][j]));

    }

    return dp[i][ignore]=ret;

}



void solve(){

    cin>>n;

    reset(dp,-1);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    lp(i,1,n+1)cin>>arr[i][0]>>arr[i][1]>>arr[i][2];

    int ans1=clac(2,0,arr[1][0]);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    int ans2=clac(2,1,arr[1][1]);

    int ans3=clac(2,2,arr[1][2]);

    //cout<<ans1<<tab<<ans2<<tab<<ans3<<endl;

    //int ans=max(ans1,(max(ans2,ans3)));

   // bug(ans1,1);

    cout<<max(ans1,(max(ans2,ans3)))<<"\n";

}



int main(){

    //FasterIO;

    #ifndef ONLINE_JUDGE

       //freopen("1.txt","r",stdin);

    #endif

    // int t;

    //cin>>t;

    //while(t--)

        solve();

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

using namespace std;



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



int n;

int arr[3][N];



int mem[N][4];

int solve(int i, int last)

{

    if(i==n)

        return 0;

    if(mem[i][last]!=-1)

        return mem[i][last];

    int op1= (last!=1) ? solve(i+1, 1)+arr[0][i] : 0;

    int op2= (last!=2) ? solve(i+1, 2)+arr[1][i] : 0;

    int op3= (last!=3) ? solve(i+1, 3)+arr[2][i] : 0;

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

}





int main()

{

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

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

    scanf("%d",&n);

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

        scanf("%d %d %d",arr[0]+i,arr[1]+i,arr[2]+i);

    memset(mem, -1, sizeof mem);

    printf("%d\n",solve(0,0));







    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.1725875763028277x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

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

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

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

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

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

#define     Endl            "\n";

#define     tab             "\t"

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

#define     sf(a)           scanf(" %d", &a);

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

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz][3],dp[sz][3];



int clac(int i,int ignore ,int value){

    if(i>n)return value;

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

    int ret=0;

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

        if(j==ignore)

            continue;

        ret=max(ret,value+clac(i+1,j,arr[i][j]));

    }

    return dp[i][ignore]=ret;

}



void solve(){

    cin>>n;

    reset(dp,-1);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    lp(i,1,n+1)cin>>arr[i][0]>>arr[i][1]>>arr[i][2];

    int ans1=clac(2,0,arr[1][0]);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    int ans2=clac(2,1,arr[1][1]);

    int ans3=clac(2,2,arr[1][2]);

    //cout<<ans1<<tab<<ans2<<tab<<ans3<<endl;

    //int ans=max(ans1,(max(ans2,ans3)));

   // bug(ans1,1);

    cout<<max(ans1,(max(ans2,ans3)))<<"\n";

}



int main(){

    //FasterIO;

    #ifndef ONLINE_JUDGE

       //freopen("1.txt","r",stdin);

    #endif

    // int t;

    //cin>>t;

    //while(t--)

        solve();

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



using namespace std;



const int N = 1e5 + 9, OO = 0x3f3f3f3f;



int n, points[N][3], dp[N][4];



int solve(int i, int prev){

    if(i == n)  return 0;



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

    if(~ret)    return ret;



    ret = 0;

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

        if(prev != j)   ret = max(ret, points[i][j] + solve(i + 1, j));





    return ret;

}



int main(){



    cin >> n;

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

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



    memset(dp, -1, sizeof dp);



    cout << solve(0, 3);

}


// Faster C++ solution for the same problem (measured speedup: 1.1794686500703941x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

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

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

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

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

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

#define     Endl            "\n";

#define     tab             "\t"

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

#define     sf(a)           scanf(" %d", &a);

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

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz][3],dp[sz][3];



int clac(int i,int ignore ,int value){

    if(i>n)return value;

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

    int ret=0;

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

        if(j==ignore)

            continue;

        ret=max(ret,value+clac(i+1,j,arr[i][j]));

    }

    return dp[i][ignore]=ret;

}



void solve(){

    cin>>n;

    reset(dp,-1);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    lp(i,1,n+1)cin>>arr[i][0]>>arr[i][1]>>arr[i][2];

    int ans1=clac(2,0,arr[1][0]);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    int ans2=clac(2,1,arr[1][1]);

    int ans3=clac(2,2,arr[1][2]);

    //cout<<ans1<<tab<<ans2<<tab<<ans3<<endl;

    //int ans=max(ans1,(max(ans2,ans3)));

   // bug(ans1,1);

    cout<<max(ans1,(max(ans2,ans3)))<<"\n";

}



int main(){

    //FasterIO;

    #ifndef ONLINE_JUDGE

       //freopen("1.txt","r",stdin);

    #endif

    // int t;

    //cin>>t;

    //while(t--)

        solve();

}