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

using namespace std;

int a[100000+5];

int b[100000+5];

int c[100000+5];



int main(int argc, char const *argv[])

{

	int N;

	while(cin >> N) {

		memset(a, 0, sizeof(a));

		memset(b, 0, sizeof(b));

		memset(c, 0, sizeof(c));

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

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

		}

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

			int tempa = a[i - 1];

			int tempb = b[i - 1];

			int tempc = c[i - 1];

			a[i] = a[i] + max(tempb, tempc);

			b[i] = b[i] + max(tempa, tempc);

			c[i] = c[i] + max(tempa, tempb);

		}

	    cout << max(a[N], max(b[N], c[N])) << endl;

    }

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.1751633057116617x):
/* BISMILLAHIR RAHMANIR RAHIM */

#include<bits/stdc++.h>

using namespace std;

#define ll        long long

#define ull       unsigned long long

#define sd(x)     scanf("%lf", &x)

#define l0(i,n)   for(ll i=0; i<n; i++)

#define MAX       100005

#define maxn 	  200005

#define pb 		  push_back

#define mk 		  make_pair

#define endl	  "\n"

#define INF 	  1000000000

#define MOD		  1000000007

#define imie(...) "[" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "





int main()

{

	int n;

	cin >> n;

	vector<int> hap[n+5];

	vector<int> dp[n+5];

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

	{

		int a,b,c;

		cin >> a >> b >> c;

		hap[i].pb(a);

		hap[i].pb(b);

		hap[i].pb(c);

		dp[i].pb(0);

		dp[i].pb(0);

		dp[i].pb(0);

	}

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

	dp[0][0] = hap[0][0];

	dp[0][1] = hap[0][1];

	dp[0][2] = hap[0][2];

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

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

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

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

	}

	int ans = *max_element(dp[n-1].begin(),dp[n-1].end());

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

//	{

//		ans = max(ans,dp[n-1][i]);

//	}

	cout << ans << endl;

	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: 2.9695371234190775x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



return 0;

}
// 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: 3.5900260880106587x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



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: 2.972331086461687x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



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: 2.1713931690145305x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



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: 2.5760954403124865x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

    cout<< max(dp[n-1][0],max(dp[n-1][1],dp[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: 2.1825991645306213x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



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: 2.1825771254676134x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



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: 2.557798197482523x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



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: 2.184402682691706x):
#include<bits/stdc++.h>



#define  ll         long long

#define  llu        unsigned long long

#define  mod        1000000007

#define  pi         ((2.0)*acos(0.0))

#define  pb         push_back

#define  pf         push_front

#define  eb         emplace_back

#define  mp         make_pair

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

#define  X          first

#define  Y          second



/*loop*/

#define  fr(i,m,n)   for(ll i=m;i<n;++i)



/*output shortcuts*/

#define  nl         "\n"

#define  Yes        cout<<"Yes\n"

#define  yes        cout<<"yes\n"

#define  YES        cout<<"YES\n"

#define  No         cout<<"No\n"

#define  no         cout<<"no\n"

#define  NO         cout<<"NO\n"



using namespace std;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    ll n;

    cin>>n;

    ll a[n][3];



    fr(i,0,n)

        fr(j,0,3)

            cin>>a[i][j];



    //ll r=1;

    ll dp[n][3]={-1};



    dp[0][0]=a[0][0];

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];



    fr(r,1,n)

    {

    for(ll i=0;i<3 && r<n; i++)  // cur_row

    {

        for(ll j=0;j<3;j++) // pre_row

        {

            if(j==i)

                continue;

            dp[r][i]=max(dp[r][i], dp[r-1][j]+a[r][i] );

        }

    }

    }



    /*cout<<nl<<nl;

    fr(i,0,n)

        {

            fr(j,0,3)

            {cout<<dp[i][j]<<" ";}

            cout<<nl;

        }

    */

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



return 0;

}