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



using namespace std;



int a[100007], b[100007], c[100007];

int memo[100007][4];

int n; 



int solve(int p, int last) {

    if(p == n) return 0;

    int & st = memo[p][last];

    if(st == -1) {

        st = max({

            last != 1? solve(p+1, 1) + a[p] : 0,

            last != 2? solve(p+1, 2) + b[p] : 0,

            last != 3? solve(p+1, 3) + c[p] : 0

        });

    }

    return st;

}

int main() {

    memset(memo, -1, sizeof memo);

    cin >> n;

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

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

    }



    cout << solve(0, 0);

}

// Faster C++ solution for the same problem (measured speedup: 2.572392368662973x):
#include "bits/stdc++.h"



#define ll long long

#define ull unsigned long long

#define ld long double



#define tiii tuple < int, int, int >

#define pii pair < int, int >

#define pll pair < ll, ll >

#define pci pair < char, int >

#define psi pair < string, int >

#define pis pair < int, string >



#define vi vector < int >

#define vd vector < double >

#define vld vector < ld >

#define vll vector < ll >

#define vs vector < string >

#define vc vector < char >

#define vpi vector < pii >

#define vpll vector < pll >

#define vpis vector < pis >



#define pb push_back

#define mt make_tuple

#define gt0 get < 0 >

#define gt1 get < 1 >

#define gt2 get < 2 >



#define pi acos(-1)



#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)



using namespace std;



const int maxn = 1e5 + 5;

const int mod = 1e9 + 9;



ll a[maxn], b[maxn], c[maxn], f[maxn][5];



int main()

{

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

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



    fast;



    int n;



    cin >> n;



    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];



    f[1][1] = a[1];

    f[1][2] = b[1];

    f[1][3] = c[1];



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

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

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

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

    }



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



    return 0;

}
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

#define format(n) fixed<<setprecision(n)

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

#define clr(dp,n) memset(dp,n,sizeof dp)

#define ex(n) return cout<<n<<finl,0;

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

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

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

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





}

int n;

vector<vector<ll>>v;

ll mem[(int)1e5 + 10][5];

ll solve(int i, int c)

{

	if (i == n)return 0;

	ll& ret = mem[i][c];

	if (~ret)return ret;

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

	{

		if (k == c)continue;

		ret = max(ret, solve(i+1, k) + v[i][k]);

	}

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		cin >> n;

	v = vector<vector<ll>>(n + 1, vector<ll>(3));

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

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

	clr(mem, -1);

	ex(solve(0, 4));

}

// Faster C++ solution for the same problem (measured speedup: 5.033638533211225x):
#include "bits/stdc++.h"



#define ll long long

#define ull unsigned long long

#define ld long double



#define tiii tuple < int, int, int >

#define pii pair < int, int >

#define pll pair < ll, ll >

#define pci pair < char, int >

#define psi pair < string, int >

#define pis pair < int, string >



#define vi vector < int >

#define vd vector < double >

#define vld vector < ld >

#define vll vector < ll >

#define vs vector < string >

#define vc vector < char >

#define vpi vector < pii >

#define vpll vector < pll >

#define vpis vector < pis >



#define pb push_back

#define mt make_tuple

#define gt0 get < 0 >

#define gt1 get < 1 >

#define gt2 get < 2 >



#define pi acos(-1)



#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)



using namespace std;



const int maxn = 1e5 + 5;

const int mod = 1e9 + 9;



ll a[maxn], b[maxn], c[maxn], f[maxn][5];



int main()

{

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

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



    fast;



    int n;



    cin >> n;



    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];



    f[1][1] = a[1];

    f[1][2] = b[1];

    f[1][3] = c[1];



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

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

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

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

    }



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



    return 0;

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

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

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

int dp[MXN][4];



int Dp(int d, int lst) {

    if(d == n) return 0;

    if(dp[d][lst] != -1) return dp[d][lst];

    int ans = 0;

    if(lst != 0) ans = Dp(d + 1, 0) + a[d];

    if(lst != 1) ans = max(ans, Dp(d + 1, 1) + b[d]);

    if(lst != 2) ans = max(ans, Dp(d + 1, 2) + c[d]);

    return dp[d][lst] = ans;

}



int main() {

    ios_base::sync_with_stdio(false); cin.tie(NULL);

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

    cout << Dp(0, 3) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.580416017642708x):
#include "bits/stdc++.h"



#define ll long long

#define ull unsigned long long

#define ld long double



#define tiii tuple < int, int, int >

#define pii pair < int, int >

#define pll pair < ll, ll >

#define pci pair < char, int >

#define psi pair < string, int >

#define pis pair < int, string >



#define vi vector < int >

#define vd vector < double >

#define vld vector < ld >

#define vll vector < ll >

#define vs vector < string >

#define vc vector < char >

#define vpi vector < pii >

#define vpll vector < pll >

#define vpis vector < pis >



#define pb push_back

#define mt make_tuple

#define gt0 get < 0 >

#define gt1 get < 1 >

#define gt2 get < 2 >



#define pi acos(-1)



#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)



using namespace std;



const int maxn = 1e5 + 5;

const int mod = 1e9 + 9;



ll a[maxn], b[maxn], c[maxn], f[maxn][5];



int main()

{

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

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



    fast;



    int n;



    cin >> n;



    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];



    f[1][1] = a[1];

    f[1][2] = b[1];

    f[1][3] = c[1];



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

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

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

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

    }



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



    return 0;

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



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

    if(dp[y][x]!=-1)

        return dp[y][x];

    if(y == 0 )

    {

    	dp[y][x] = A[y][x];

        return A[y][x];

    }

    return dp[y][x] = max(solver((y-1),(x+1)%3)+ A[y][x],

	solver((y-1),(x+2)%3) + A[y][x]);

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.176209966716951x):
#include "bits/stdc++.h"



#define ll long long

#define ull unsigned long long

#define ld long double



#define tiii tuple < int, int, int >

#define pii pair < int, int >

#define pll pair < ll, ll >

#define pci pair < char, int >

#define psi pair < string, int >

#define pis pair < int, string >



#define vi vector < int >

#define vd vector < double >

#define vld vector < ld >

#define vll vector < ll >

#define vs vector < string >

#define vc vector < char >

#define vpi vector < pii >

#define vpll vector < pll >

#define vpis vector < pis >



#define pb push_back

#define mt make_tuple

#define gt0 get < 0 >

#define gt1 get < 1 >

#define gt2 get < 2 >



#define pi acos(-1)



#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)



using namespace std;



const int maxn = 1e5 + 5;

const int mod = 1e9 + 9;



ll a[maxn], b[maxn], c[maxn], f[maxn][5];



int main()

{

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

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



    fast;



    int n;



    cin >> n;



    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];



    f[1][1] = a[1];

    f[1][2] = b[1];

    f[1][3] = c[1];



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

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

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

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

    }



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



    return 0;

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

 

#define pb push_back

#define fst first

#define snd second

#define INF 0x7fffffff

#define INFLL 0x7FFFFFFFFFFFFFFF

#define lenA(ar) sizeof(ar)/sizeof(*ar)

#define lenrows(mat) sizeof(mat)/sizeof(*mat)

#define lencols(mat) sizeof(*mat)/sizeof(**mat)

#define fill(a,c) memset(&a, c, sizeof(a))

#define fore(i,x,y) for(ll i=x;i<y;i++)

#define foreI(i,x,y) for(ll i=x-1;i>=y;i--)

#define max2(a,b) max(a, b)

#define max3(a,b,c) max2(max2(a,b),c)

#define min2(a,b) min(a, b)

#define min3(a,b,c) min2(min2(a,b),c)

#define MOD 1000000009

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

#define rall(x) (x).rbegin(), (x).rend()

#define FIN ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

 

typedef long long ll;

typedef unsigned long long ull;

 

using namespace std;

ll arr[100010][3];

ll dp[100010][3];

int n;

ll solve(int i, int j){

    if(i==n-1) return arr[i][j];

    ll &r = dp[i][j];

    if(r>-1) return r;

    ll a = 0; ll b = 0;ll c = 0; ll act=arr[i][j];

    if(j!=0) a=solve(i+1, 0)+act;

    if(j!=1) b=solve(i+1, 1)+act;

    if(j!=2) c=solve(i+1, 2)+act;

   return r=max3(a,b,c);

}

 

int main() {

     

    cin>>n;

    arr[n][3];

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

    fore(i,0,n){

        fore(j,0,3){

            cin>>arr[i][j];

        }

    }

 

    cout<<max3(solve(0,0),solve(0,1),solve(0,2))<<"\n";

 

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.359087617428655x):
#include "bits/stdc++.h"



#define ll long long

#define ull unsigned long long

#define ld long double



#define tiii tuple < int, int, int >

#define pii pair < int, int >

#define pll pair < ll, ll >

#define pci pair < char, int >

#define psi pair < string, int >

#define pis pair < int, string >



#define vi vector < int >

#define vd vector < double >

#define vld vector < ld >

#define vll vector < ll >

#define vs vector < string >

#define vc vector < char >

#define vpi vector < pii >

#define vpll vector < pll >

#define vpis vector < pis >



#define pb push_back

#define mt make_tuple

#define gt0 get < 0 >

#define gt1 get < 1 >

#define gt2 get < 2 >



#define pi acos(-1)



#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)



using namespace std;



const int maxn = 1e5 + 5;

const int mod = 1e9 + 9;



ll a[maxn], b[maxn], c[maxn], f[maxn][5];



int main()

{

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

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



    fast;



    int n;



    cin >> n;



    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];



    f[1][1] = a[1];

    f[1][2] = b[1];

    f[1][3] = c[1];



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

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

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

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

    }



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



    return 0;

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

#include <iostream>

using namespace std;

const int MAXN=1e5+7;

int n,ans;

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

int main()

{

	cin>>n;

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

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

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

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

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

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

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

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

			{

				int maxx=0;

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

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

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

			}

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

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

	cout<<ans<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 11.36609820047917x):
#include "bits/stdc++.h"



#define ll long long

#define ull unsigned long long

#define ld long double



#define tiii tuple < int, int, int >

#define pii pair < int, int >

#define pll pair < ll, ll >

#define pci pair < char, int >

#define psi pair < string, int >

#define pis pair < int, string >



#define vi vector < int >

#define vd vector < double >

#define vld vector < ld >

#define vll vector < ll >

#define vs vector < string >

#define vc vector < char >

#define vpi vector < pii >

#define vpll vector < pll >

#define vpis vector < pis >



#define pb push_back

#define mt make_tuple

#define gt0 get < 0 >

#define gt1 get < 1 >

#define gt2 get < 2 >



#define pi acos(-1)



#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)



using namespace std;



const int maxn = 1e5 + 5;

const int mod = 1e9 + 9;



ll a[maxn], b[maxn], c[maxn], f[maxn][5];



int main()

{

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

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



    fast;



    int n;



    cin >> n;



    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];



    f[1][1] = a[1];

    f[1][2] = b[1];

    f[1][3] = c[1];



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

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

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

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

    }



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



    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: 1.7001867258130499x):
#include <bits/stdc++.h>

using namespace std;

#define sz 100005

int n,a[sz],b[sz],c[sz];

int dp[sz][5];



int res(int ind,int ofStatus=0)

{

    if(ind==n)return 0;

    if(dp[ind][ofStatus]!=-1)return dp[ind][ofStatus];

    int r=INT_MIN;

    if(ofStatus!=1)

    {

        r=max(r,a[ind]+res(ind+1,1));

    }

    if(ofStatus!=2)

    {

        r=max(r,b[ind]+res(ind+1,2));

    }

    if(ofStatus!=3)

    {

        r=max(r,c[ind]+res(ind+1,3));

    }

    return dp[ind][ofStatus]=r;

}



int main()

{

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

    cin>>n;

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

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



    cout<<res(0)<<endl;



   return 0;

}
// Slower C++ solution:
/*

** author:mahmoud magdy

*/

#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define endl "\n"



inline int D()

{

    int m ;

    cin>>m;

    return m ;

}

inline ll lD()

{

    ll m ;

    cin>>m;

    return m ;

}

inline double dD()

{

    double m;

    cin>>m;

    return m;

}

void fast()

{

    ios_base :: sync_with_stdio( 0 ) ;

    cin.tie( 0 ) ;

    cout.tie( 0 ) ;

}

int n;

ll a[100005],b[100005],c[100005];

ll dp[100005][5];

ll solve(int i,int prv){

 if(i==n)return 0;

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

 ll x=0;

 if(prv!=1){

 x=a[i]+solve(i+1,1);

 }

 ll y=0;

 if(prv!=2)

 y=b[i]+solve(i+1,2);

 ll z=0;

 if(prv!=3)

 z=c[i]+solve(i+1,3);

 return dp[i][prv]=max({x,y,z});

}

int main()

{

    memset(dp,-1,sizeof dp);

    fast();

    n=D();

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

        a[i]=D();b[i]=D();c[i]=D();

    }

    cout<<solve(0,10)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.6970263427736925x):
#include <bits/stdc++.h>

using namespace std;

#define sz 100005

int n,a[sz],b[sz],c[sz];

int dp[sz][5];



int res(int ind,int ofStatus=0)

{

    if(ind==n)return 0;

    if(dp[ind][ofStatus]!=-1)return dp[ind][ofStatus];

    int r=INT_MIN;

    if(ofStatus!=1)

    {

        r=max(r,a[ind]+res(ind+1,1));

    }

    if(ofStatus!=2)

    {

        r=max(r,b[ind]+res(ind+1,2));

    }

    if(ofStatus!=3)

    {

        r=max(r,c[ind]+res(ind+1,3));

    }

    return dp[ind][ofStatus]=r;

}



int main()

{

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

    cin>>n;

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

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



    cout<<res(0)<<endl;



   return 0;

}
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

#define format(n) fixed<<setprecision(n)

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

#define clr(dp,n) memset(dp,n,sizeof dp)

#define ex(n) return cout<<n<<finl,0;

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

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

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

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





}

int n;

vector<vector<ll>>v;

ll mem[(int)1e5 + 10][5];

ll solve(int i, int c)

{

	if (i == n)return 0;

	ll& ret = mem[i][c];

	if (~ret)return ret;

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

	{

		if (k == c)continue;

		ret = max(ret, solve(i+1, k) + v[i][k]);

	}

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		cin >> n;

	v = vector<vector<ll>>(n + 1, vector<ll>(3));

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

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

	clr(mem, -1);

	ex(solve(0, 4));

}

// Faster C++ solution for the same problem (measured speedup: 1.6980633178024438x):
#include <bits/stdc++.h>

using namespace std;

#define sz 100005

int n,a[sz],b[sz],c[sz];

int dp[sz][5];



int res(int ind,int ofStatus=0)

{

    if(ind==n)return 0;

    if(dp[ind][ofStatus]!=-1)return dp[ind][ofStatus];

    int r=INT_MIN;

    if(ofStatus!=1)

    {

        r=max(r,a[ind]+res(ind+1,1));

    }

    if(ofStatus!=2)

    {

        r=max(r,b[ind]+res(ind+1,2));

    }

    if(ofStatus!=3)

    {

        r=max(r,c[ind]+res(ind+1,3));

    }

    return dp[ind][ofStatus]=r;

}



int main()

{

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

    cin>>n;

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

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



    cout<<res(0)<<endl;



   return 0;

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

 

#define pb push_back

#define fst first

#define snd second

#define INF 0x7fffffff

#define INFLL 0x7FFFFFFFFFFFFFFF

#define lenA(ar) sizeof(ar)/sizeof(*ar)

#define lenrows(mat) sizeof(mat)/sizeof(*mat)

#define lencols(mat) sizeof(*mat)/sizeof(**mat)

#define fill(a,c) memset(&a, c, sizeof(a))

#define fore(i,x,y) for(ll i=x;i<y;i++)

#define foreI(i,x,y) for(ll i=x-1;i>=y;i--)

#define max2(a,b) max(a, b)

#define max3(a,b,c) max2(max2(a,b),c)

#define min2(a,b) min(a, b)

#define min3(a,b,c) min2(min2(a,b),c)

#define MOD 1000000009

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

#define rall(x) (x).rbegin(), (x).rend()

#define FIN ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

 

typedef long long ll;

typedef unsigned long long ull;

 

using namespace std;

ll arr[100010][3];

ll dp[100010][3];

int n;

ll solve(int i, int j){

    if(i==n-1) return arr[i][j];

    ll &r = dp[i][j];

    if(r>-1) return r;

    ll a = 0; ll b = 0;ll c = 0; ll act=arr[i][j];

    if(j!=0) a=solve(i+1, 0)+act;

    if(j!=1) b=solve(i+1, 1)+act;

    if(j!=2) c=solve(i+1, 2)+act;

   return r=max3(a,b,c);

}

 

int main() {

     

    cin>>n;

    arr[n][3];

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

    fore(i,0,n){

        fore(j,0,3){

            cin>>arr[i][j];

        }

    }

 

    cout<<max3(solve(0,0),solve(0,1),solve(0,2))<<"\n";

 

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1331650905813375x):
#include <bits/stdc++.h>

using namespace std;

#define sz 100005

int n,a[sz],b[sz],c[sz];

int dp[sz][5];



int res(int ind,int ofStatus=0)

{

    if(ind==n)return 0;

    if(dp[ind][ofStatus]!=-1)return dp[ind][ofStatus];

    int r=INT_MIN;

    if(ofStatus!=1)

    {

        r=max(r,a[ind]+res(ind+1,1));

    }

    if(ofStatus!=2)

    {

        r=max(r,b[ind]+res(ind+1,2));

    }

    if(ofStatus!=3)

    {

        r=max(r,c[ind]+res(ind+1,3));

    }

    return dp[ind][ofStatus]=r;

}



int main()

{

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

    cin>>n;

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

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



    cout<<res(0)<<endl;



   return 0;

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

#include <iostream>

using namespace std;

const int MAXN=1e5+7;

int n,ans;

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

int main()

{

	cin>>n;

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

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

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

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

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

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

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

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

			{

				int maxx=0;

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

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

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

			}

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

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

	cout<<ans<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.834275006723879x):
#include <bits/stdc++.h>

using namespace std;

#define sz 100005

int n,a[sz],b[sz],c[sz];

int dp[sz][5];



int res(int ind,int ofStatus=0)

{

    if(ind==n)return 0;

    if(dp[ind][ofStatus]!=-1)return dp[ind][ofStatus];

    int r=INT_MIN;

    if(ofStatus!=1)

    {

        r=max(r,a[ind]+res(ind+1,1));

    }

    if(ofStatus!=2)

    {

        r=max(r,b[ind]+res(ind+1,2));

    }

    if(ofStatus!=3)

    {

        r=max(r,c[ind]+res(ind+1,3));

    }

    return dp[ind][ofStatus]=r;

}



int main()

{

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

    cin>>n;

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

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



    cout<<res(0)<<endl;



   return 0;

}