stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.5609454277263133x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

#define fast ios_base::sync_with_stdio(false),cin.tie(NULL)

#define x first

#define y second



vector<vector<int>> dp, arr;

int main() {

    fast;

    int n;

    cin>>n;

    arr.resize(n);

    dp.resize(n);

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

        arr[i].resize(3);

        dp[i].resize(3);

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

            cin>>arr[i][j];

            dp[i][j] = -1;

        }

    }

    for (int i = n - 1; i >= 0; i--) {

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

            if(i == n-1)

            {

                dp[i][j] = arr[i][j];

            } else{

                int temp, max = 0;

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

                    if(k == j)

                        continue;

                    temp = dp[i + 1][k];

                    if(temp > max)

                        max = temp;

                }

                dp[i][j] = max + arr[i][j];

            }

        }

    }

    sort(dp[0].begin(), dp[0].end());

    cout<<dp[0][2];

}
// 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: 2.967833033308671x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 3.5879659258714374x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 2.9706253930156117x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 2.170147099519073x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 7.548178730510557x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 2.574617129525031x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 2.181346664394514x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 2.1813246379787796x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 2.55633038669869x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 2.1831491475938067x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}
// 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: 2.5713316525492185x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

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

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

            if (j==0) dp[i][j]= a[i] + max(dp[i-1][1], dp[i-1][2]);

            if (j==1) dp[i][j]= b[i] + max(dp[i-1][0], dp[i-1][2]);

            if (j==2) dp[i][j]= c[i] + max(dp[i-1][0], dp[i-1][1]);

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}