stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <cstring>

#include <iostream>

using namespace std;

const int MAXN=1e5+7;

int n,ans;

int h[MAXN][5],f[MAXN][5][5];

int main()

{

	cin>>n;

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

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

	memset(f,127,sizeof(f));

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

		for(int j=1;j<=3;j++) f[1][i][j]=h[1][i];

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

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

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

			{

				int maxx=0;

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

					if(j!=k&&k!=l) maxx=max(maxx,f[i-1][k][l]);

				f[i][j][k]=maxx+h[i][j];

			}

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

		for(int j=1;j<=3;j++) ans=max(ans,f[n][i][j]);

	cout<<ans<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3534356408696684x):
#include <bits/stdc++.h>

#define ll long long 

#define vpnt(ans) for(ll i = 0; i < ans.size(); i++) cout << ans[i] << (i + 1 < ans.size() ? ' ' : '\n');

#define setbit(x,k) (x |= (1 << k))

#define clearbit(x,k) (x &= ~(1ll << k))

#define checkbit(x,k) (x & (1ll << k))

#define mp make_pair

#define scl(x) scanf("%lld",&x)

#define sci(x) scanf("%d",&x)

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define pii pair<int,int> 

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

#define nn printf("\n") 

#define mod 10000007

using namespace std;    



ll n, a[100005][3];

ll dp[100005][3][3];



ll solve(int pos, int konta, int koydin)

{

    if(pos == n) return 0;

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



    ll res = 0;

    if(koydin<1) res = max(res, a[pos][konta] + solve(pos+1, konta, koydin+1));

    if(konta == 0)

    {

        res = max(res, a[pos][1] + solve(pos+1, 1, 1));

        res = max(res, a[pos][2] + solve(pos+1, 2, 1));

    }

    if(konta == 1)

    {

        res = max(res, a[pos][0] + solve(pos+1, 0, 1));

        res = max(res, a[pos][2] + solve(pos+1, 2, 1));

    }

    if(konta == 2)

    {

        res = max(res, a[pos][1] + solve(pos+1, 1, 1));

        res = max(res, a[pos][0] + solve(pos+1, 0, 1));

    }

    return dp[pos][konta][koydin] = res;

}



int main()

{

    cin>>n;

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

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

    {

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

    }

    cout<<solve(0,2,0); nn;

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

#include <algorithm>

#include <cstring>

const int maxn = 3e5 + 5;

using namespace std;



int N;

int a[maxn][4];

int dp[maxn][4];

int main()

{

	scanf("%d", &N);

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

	{

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

		{

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

		}

	}

	memset(dp, 0, sizeof dp);

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

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

	for(int i = 2;i <= N;i++)

	{

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

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

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

	}

	int ans = 0;

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

		ans = max(dp[N][i], ans);

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

	return 0;	

}

// Faster C++ solution for the same problem (measured speedup: 5.037783341537549x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



LL dp1[N],dp2[N],dp3[N];



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

        dp1[i] = one[i]+max(dp2[i-1],dp3[i-1]);

        dp2[i] = two[i]+max(dp1[i-1],dp3[i-1]);

        dp3[i] = three[i]+max(dp1[i-1],dp2[i-1]);

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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

using namespace std;

#define db(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<'\n'

const int N = 1e5;

int A[N+7][5], mem[N+7][5];



int dp(int i, int j) {

	if(i==0) return 0;

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

	if(j==1) return mem[i][j] = A[i][j] + max(dp(i-1, 2), dp(i-1, 3));

	if(j==2) return mem[i][j] = A[i][j] + max(dp(i-1, 3), dp(i-1, 1));

	else return mem[i][j] = A[i][j] + max(dp(i-1, 1), dp(i-1, 2));

}

int main(){

	ios_base::sync_with_stdio(0);

	int a,b,n,T,cas=0;

	cin>>n;

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

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

	}

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

	int res = max(dp(n,1), max(dp(n,2), dp(n,3)));

	cout<<res<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.972379358789273x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



LL dp1[N],dp2[N],dp3[N];



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

        dp1[i] = one[i]+max(dp2[i-1],dp3[i-1]);

        dp2[i] = two[i]+max(dp1[i-1],dp3[i-1]);

        dp3[i] = three[i]+max(dp1[i-1],dp2[i-1]);

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

}
// Slower C++ solution:
/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



int n;

int a[N][3];

int mem[N][3];



int go(int idx , int lst){

    if(idx == n + 1)

        return 0;

    if(~mem[idx][lst])

        return mem[idx][lst];

    int best = 0;

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

        if(idx == 1 || lst != i){

            best = max(best , go(idx + 1 , i) + a[idx][i]);

        }

    }

    return mem[idx][lst] = best;

}



int main(){

    scanf("%d" , &n);

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

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

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

        }

    }

    memset(mem , -1 , sizeof mem);

    printf("%d\n" , go(1 , 0));

}


// Faster C++ solution for the same problem (measured speedup: 2.1574085116207433x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



LL dp1[N],dp2[N],dp3[N];



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

        dp1[i] = one[i]+max(dp2[i-1],dp3[i-1]);

        dp2[i] = two[i]+max(dp1[i-1],dp3[i-1]);

        dp3[i] = three[i]+max(dp1[i-1],dp2[i-1]);

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

}
// 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: 2.17575568432744x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



LL dp1[N],dp2[N],dp3[N];



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

        dp1[i] = one[i]+max(dp2[i-1],dp3[i-1]);

        dp2[i] = two[i]+max(dp1[i-1],dp3[i-1]);

        dp3[i] = three[i]+max(dp1[i-1],dp2[i-1]);

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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

using namespace std;



int m;

const int mx = 1e5+5;

int tasks[mx][3];

int diario[3][mx];

int cosa(int a, int i){

	if(i == m){

     return 0;

     }



	if(diario[a][i] != -1) {

        return diario[a][i];

    }

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



        if(a==0){

		ans=max(ans, cosa(1, i+1)+tasks[i][0]);

		ans=max(ans, cosa(2, i+1)+tasks[i][0]);

	}

        if(a==1){

		ans=max(ans, cosa(0, i+1)+tasks[i][1]);

		ans=max(ans, cosa(2, i+1)+tasks[i][1]);

        }

	if(a==2){

		ans=max(ans, cosa(0, i+1)+tasks[i][2]);

		ans=max(ans, cosa(1, i+1)+tasks[i][2]);

	}



	return ans;

}

int main(){

	cin>>m;

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

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

	}

	memset(diario, -1, sizeof diario);

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.175781613713621x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



LL dp1[N],dp2[N],dp3[N];



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

        dp1[i] = one[i]+max(dp2[i-1],dp3[i-1]);

        dp2[i] = two[i]+max(dp1[i-1],dp3[i-1]);

        dp3[i] = three[i]+max(dp1[i-1],dp2[i-1]);

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

}
// 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;



int nn;

const int nxx = 1e5+5;

int acc[nxx][3];

int dpp[3][nxx];

int go(int aa, int ii)

{

	if(ii==nn) return 0;

	if(dpp[aa][ii] != -1) return dpp[aa][ii];

	int& ans = dpp[aa][ii];

	if(aa==0)

	{

		ans=max(ans, go(1, ii+1)+acc[ii][0]);

		ans=max(ans, go(2, ii+1)+acc[ii][0]);

	}

	if(aa==1)

	{

		ans=max(ans, go(0, ii+1)+acc[ii][1]);

		ans=max(ans, go(2, ii+1)+acc[ii][1]);

	}

	if(aa==2)

	{

		ans=max(ans, go(0, ii+1)+acc[ii][2]);

		ans=max(ans, go(1, ii+1)+acc[ii][2]);

	}



	return ans;



}

int main()

{

    cin>>nn;

    for(int ii=0; ii<nn; ii++) cin>>acc[ii][0]>>acc[ii][1]>>acc[ii][2];

    memset(dpp, -1, sizeof dpp);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.175779428668275x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



LL dp1[N],dp2[N],dp3[N];



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

        dp1[i] = one[i]+max(dp2[i-1],dp3[i-1]);

        dp2[i] = two[i]+max(dp1[i-1],dp3[i-1]);

        dp3[i] = three[i]+max(dp1[i-1],dp2[i-1]);

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

}
// 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;



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: 2.17575568432744x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



LL dp1[N],dp2[N],dp3[N];



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

        dp1[i] = one[i]+max(dp2[i-1],dp3[i-1]);

        dp2[i] = two[i]+max(dp1[i-1],dp3[i-1]);

        dp3[i] = three[i]+max(dp1[i-1],dp2[i-1]);

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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

using namespace std;

int v[100009][3];

int dp[100009][3];

int n;

int vacation(int i, int j)

{

    if(i==n-1)

        return dp[i][j]=v[i][j];

    if(i>=n || j>=3)

        return 0;

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

        return dp[i][j];

    int temp = INT_MIN;

        if(j==0)

        {

             temp = max(vacation(i+1,j+1)+v[i][j],temp);

        temp = max(vacation(i+1,j+2)+v[i][j],temp);

 return dp[i][j]= temp;

        }

      else  if(j==1)

        {

             temp = max(vacation(i+1,0)+v[i][j],temp);

        temp = max(vacation(i+1,2)+v[i][j],temp);

 return dp[i][j]= temp;

        }

        else

        {

             temp = max(vacation(i+1,1)+v[i][j],temp);

        temp = max(vacation(i+1,0)+v[i][j],temp);

         return dp[i][j]= temp;

        }

}



int main()

{

    cin>>n;

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

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

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

    cout<<max(vacation(0,0),max(vacation(0,1),vacation(0,2)))<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 2.1758758808691945x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



LL dp1[N],dp2[N],dp3[N];



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

        dp1[i] = one[i]+max(dp2[i-1],dp3[i-1]);

        dp2[i] = two[i]+max(dp1[i-1],dp3[i-1]);

        dp3[i] = three[i]+max(dp1[i-1],dp2[i-1]);

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

}