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


#include<bits/stdc++.h>

using namespace std;



#define MAX 100000



int dp[MAX+5][4];

int val[MAX+5][4];

int n;



void initialize()

{

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

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

            dp[i][j] = -1;

}



int vacation(int i, int j)

{

    if(i<=0) return 0;



  /*  if(i==1)

    {

        return val[i][j];

    }

*/

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



    int mx = INT_MIN;



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

    {

        if(j==k) continue;



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

    }





    dp[i][j] = mx;



    return mx;

}



int main()

{

    initialize();



    cin>>n;

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

    {

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

            cin>>val[i][j];

    }



    int retVal1 = vacation(n, 1);



    initialize();



    int retVal2 = vacation(n, 2);



    initialize();



    int retVal3 = vacation(n, 3);



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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.6262092807940083x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

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

using namespace std;

int arr[100010][5];

int dp[100010][5];



int vac(int a, int b){

	if(a == 0) return 0;

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

	

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

		if(i != b){

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

		}

	}

	

	return dp[a][b];

}



int main (){

	memset(dp,-1,sizeof dp);

	int n;

	cin >> n;

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

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

	}

	

	

	int ans = 0;

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

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

	}

	

	cout << ans << endl;

	

}

// Faster C++ solution for the same problem (measured speedup: 3.0022886483514344x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

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

using namespace std;

#define INF INT_MAX

#define N 100005

#define K 3

int n, k;

int arr[N][K];

int dp[N][K];

int solve(int idx, int last_k){

    int res;

    if(idx == n)

        return 0;

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

        return dp[idx][last_k];

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

        if(last_k != i){

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

        }

    }

    return dp[idx][last_k];

}

int main()

{

    cin >> n;

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

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

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

            cin >> arr[i][j];

    cout << solve(0, -1);

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 2.1932782293781x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

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

typedef long long ll;

using namespace std;

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

const double EPS = 1e-3;

#define M_PI 3.14159265358979323846

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

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

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

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

#define pb push_back

#define F first

#define S second

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

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

typedef pair <int,int> pii ;

typedef vector < pii > vii ;

typedef vector < int > vi;

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

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

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

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

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

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







int n,x;

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

ll dp[N][4];





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



    if(idx == n) return 0;



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

    if(~ret) return ret;



    ret = -OO;

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

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

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



    return ret;



}







int main(){





    //ios_base::sync_with_stdio(0);

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





    scanf("%d",&n);

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

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



    memset(dp, -1 , sizeof dp);

    ll ans = solve(0 , 0);

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







    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.628633139547289x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

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

#include <cstring>

using namespace std;

const int maxn=1e5+5;

const int inf=0x3f3f3f3f;

int dp[maxn][4];

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

int main()

{

   int n;

   cin>>n;

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

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

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

   }

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

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

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

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

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

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

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

   }

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

   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.6020594181945484x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

//const int mod = 1e9+9;

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

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

*void exEuclid(int a, int b){

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

	exEuclid(b, a%b);

	int x1=y;

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

	x=x1; 100000

	y=y1;

} 

* */

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

int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i){

	if(i == n) return 0;

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

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

	

	if(a==0){

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

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

	}

	if(a==1){

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

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

	}

	if(a==2){

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

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

	}

	

	return ans;

}

int main(){

	cin>>n;

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

	memset(dp, -1, sizeof dp);

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.2045971679999363x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

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

using namespace std;

const int oo = 1e9;



int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i)

{

	if(i==n) return 0;

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

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

	if(a==0)

	{

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

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

	}

	if(a==1)

	{

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

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

	}

	if(a==2)

	{

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

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

	}

	

	return ans;

	

}

int main()

{

    cin>>n;

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

    memset(dp, -1, sizeof dp);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2045749068093876x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

}
// 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: 2.583577760920653x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	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: 2.206418863478876x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	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: 2.598738922028289x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

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

#include <algorithm>

#include <iostream>

#include <iomanip>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <vector>

#include <string>

#include <climits>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <deque>

#include <stdio.h>

//#define int long long

using namespace std;

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

#define endl '\n'

#define OnlineJudgec

#define F first

#define S second

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

#define clr(v,d ) memset(v, d, sizeof(v))

typedef long long ll;

typedef vector<int> vi;

void init()

{

    cin.tie(0);

    cin.sync_with_stdio(0);

    cout.tie(0);

    #ifdef OnlineJudge

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

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

    #endif

}

ll S(ll x){ll s = 0;while(x){s += x % 10;x /= 10;}return s;} /// sum of digits.

ll powe(ll a , ll n){if(n == 1)return a;return a * powe(a ,n-1);}

ll com(int n,int m){ll z = 1;for(int i = 0; i < m;i++){z = z * ( n - i) / (i + 1) ;}return z;}

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

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

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

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

/// <<------------------------------------------------------------------------------>>

const int mx = 100009;

int n , k;

int arr[mx][4];

int dp[mx][4];

int solve(int idx , int last)

{

     if(idx == n)return 0;

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

     if(~ret) return ret;

     int A = 0 ,B = 0 ,C = 0;

     if(last != 0)

        A = solve(idx+1 , 0) + arr[idx][0];

     if(last != 1)

        B = solve(idx+1 , 1) + arr[idx][1];

     if(last != 2)

        C = solve(idx+1 , 2) + arr[idx][2];

     return ret = max(A,max(B,C));

}

/// <<------------------------------------------------------------------------------>>

int main()

{

    init();

    cin >> n;

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

    {

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

    }

    clr(dp,-1);

    cout<<solve(0,3);

}


// Faster C++ solution for the same problem (measured speedup: 2.613231549861721x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

}
// Slower C++ solution:
/*

██╗  ██╗ █████╗ ██████╗ ███████╗███████╗███╗   ███╗        ██╗  ██╗██╗  ██╗ █████╗ ██╗     ███████╗██████╗

██║ ██╔╝██╔══██╗██╔══██╗██╔════╝██╔════╝████╗ ████║        ██║ ██╔╝██║  ██║██╔══██╗██║     ██╔════╝██╔══██╗

█████╔╝ ███████║██████╔╝█████╗  █████╗  ██╔████╔██║        █████╔╝ ███████║███████║██║     █████╗  ██║  ██║

██╔═██╗ ██╔══██║██╔══██╗██╔══╝  ██╔══╝  ██║╚██╔╝██║        ██╔═██╗ ██╔══██║██╔══██║██║     ██╔══╝  ██║  ██║

██║  ██╗██║  ██║██║  ██║███████╗███████╗██║ ╚═╝ ██║███████╗██║  ██╗██║  ██║██║  ██║███████╗███████╗██████╔╝

╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═╝     ╚═╝╚══════╝╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═════╝

Believe in yourself ,( try and try and then try ), You can do it 

*/

#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<unordered_map>

#define endl "\n"

#define Ceil(x,y) ((x+y-1)/y)

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

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

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

#define rt(s) return cout<<s,0

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

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

#define OO 0x3f3f3f3f3f3f3f3fLL

const int oo = 0x3f3f3f3f, mod = 1e9 + 7;

using namespace std;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

//x = xDegrees*3.14159265 / 180;

//ef1tn=(( (n / 2)*( (n / 2) + 1) / 2) * 2);

//of1tn=(n * (n + 1) / 2 - 2 * ((n / 2) * (n / 2 + 1) / 2));

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

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

void K_K(){

#ifndef ONLINE_JUDGE

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

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

#endif  !ONLINEJUDGE

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

}

ll gcd(ll a, ll b) {

	return b == 0 ? abs(a) : gcd(b, a % b);

}

ll lcm(ll a, ll b) {

	return a / gcd(a, b) * b;

}

int n;

vector<int>a, b, c;

int dp[100005][5];

int sol(int idx, int pre){

	if (idx == n)

		return 0;

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

	if (~ret)

		return ret;

	int ch1 = 0, ch2 = 0, ch3 = 0;

	if (pre != 1)

		ch1 = sol(idx + 1, 1) + a[idx];

	if (pre != 2)

		ch2 = sol(idx + 1, 2) + b[idx];

	if (pre != 3)

		ch3 = sol(idx + 1, 3) + c[idx];

	return ret = max({ ch1, ch2, ch3 });

}

int main(){

	K_K();

	cin >> n;

	a.resize(n);

	b.resize(n);

	c.resize(n);

	clr(dp, -1);

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

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

	cout << sol(0, 0);

}

// Faster C++ solution for the same problem (measured speedup: 3.014873993403844x):
#include<iostream>

#include<cmath>

#include<iomanip>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

int a[100005][5],dp[100005][5];

int main()

{

    ll n,i,j,k,ans=0;

    cin>>n;

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

    {

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

    	cin>>a[i][j];

	}

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

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

	dp[1][3]=a[1][3];

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

	{

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

		{

			ll mi=0,x;

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

			{

				if(k!=j)

				{

					x=dp[i-1][k]+a[i][j];

					mi=mi>x?mi:x;

				}

			}

			dp[i][j]=mi;

		}

	}

	ans=dp[n][1]>dp[n][2]?dp[n][1]:dp[n][2];

	ans=ans>dp[n][3]?ans:dp[n][3];

	cout<<ans;

	return 0;

}