stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 5.115628649231528x):
#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   return 0;

}
// 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: 3.01830943757221x):
#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   return 0;

}
// 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: 2.190745421531893x):
#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   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: 2.2093761001394676x):
#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   return 0;

}
// 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: 2.2094024301941384x):
#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   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;



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: 2.20940021138483x):
#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   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;



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: 2.2093761001394676x):
#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   return 0;

}
// 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: 2.209498153993484x):
#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   return 0;

}
// 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: 5.10773737127483x):
#include<bits/stdc++.h>

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

#define read(a) scanf("%d",&a)

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000;



using namespace std;



int main()

{

    int n,m;

    cin>>n;

    int ar[n+5][4];

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

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

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

        }

    }

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

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

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

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

    }

    cout<<max(ar[n-1][0],max(ar[n-1][1],ar[n-1][2]));



    return 0;

}
// 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: 3.01365344700597x):
#include<bits/stdc++.h>

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

#define read(a) scanf("%d",&a)

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000;



using namespace std;



int main()

{

    int n,m;

    cin>>n;

    int ar[n+5][4];

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

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

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

        }

    }

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

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

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

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

    }

    cout<<max(ar[n-1][0],max(ar[n-1][1],ar[n-1][2]));



    return 0;

}
// 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: 2.1873660165283124x):
#include<bits/stdc++.h>

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

#define read(a) scanf("%d",&a)

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000;



using namespace std;



int main()

{

    int n,m;

    cin>>n;

    int ar[n+5][4];

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

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

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

        }

    }

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

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

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

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

    }

    cout<<max(ar[n-1][0],max(ar[n-1][1],ar[n-1][2]));



    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: 2.205967955781744x):
#include<bits/stdc++.h>

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

#define read(a) scanf("%d",&a)

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000;



using namespace std;



int main()

{

    int n,m;

    cin>>n;

    int ar[n+5][4];

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

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

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

        }

    }

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

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

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

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

    }

    cout<<max(ar[n-1][0],max(ar[n-1][1],ar[n-1][2]));



    return 0;

}
// 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: 2.2059942452201398x):
#include<bits/stdc++.h>

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

#define read(a) scanf("%d",&a)

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000;



using namespace std;



int main()

{

    int n,m;

    cin>>n;

    int ar[n+5][4];

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

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

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

        }

    }

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

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

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

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

    }

    cout<<max(ar[n-1][0],max(ar[n-1][1],ar[n-1][2]));



    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;



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: 2.2059920298335274x):
#include<bits/stdc++.h>

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

#define read(a) scanf("%d",&a)

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000;



using namespace std;



int main()

{

    int n,m;

    cin>>n;

    int ar[n+5][4];

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

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

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

        }

    }

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

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

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

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

    }

    cout<<max(ar[n-1][0],max(ar[n-1][1],ar[n-1][2]));



    return 0;

}