stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
//       Author: Mostafa Mounir Shehab

 

#include <bits/stdc++.h>

using namespace std;

 

#define PI  3.14159265  // PI = acos(-1)

#define EPS (1e-10)

#define endl "\n"

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

typedef long long ll;

typedef long double ld;

//template <class Type>

 

 

void Fast(){

	std::ios_base::sync_with_stdio(0);

	cin.tie(NULL); cout.tie(NULL); cerr.tie(NULL);

}

 

inline int D(){

	int t;

	scanf("%d",&t);

	return t;

}

inline ll llD(){

	ll t;

	scanf("%lld",&t);

	return t;

}

 

 

ll MOD=1e9 + 7; 

const int N=1e5 + 5,M=50 + 1;

int arr[N][4];

ll dp[N][4];

ll OO=1e12; 

 

ll n,m,k,cnt,SU,idx,len,MN,MX;

ll l,r,mid;

ll a,b,c,d;



ll Solve(int idx,int row){

	

		

	if(idx == n)

		return 0;

	

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

	if(~ret)

		return ret;

	

	ret = -OO;

	

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

		if(row != i)

			ret = max(ret,arr[idx][i] + Solve(idx+1,i));

	

	return ret;

}







 

 

int main(){

	

	memset(dp, -1 ,sizeof dp);

	cin>>n;

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

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



	cout<<Solve(0,0)<<endl;

	



	





}

// Faster C++ solution for the same problem (measured speedup: 4.283221777522511x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

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

#include <bits/stdc++.h>

#include <unordered_map>

#define stop(x) {cout<<x<<endl;exit(0);}

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

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

#define endl '\n'

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

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

typedef long long ll;

using namespace std;



void file() {

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

#ifndef ONLINE_JUDGE

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

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

#endif

}





ll n;

ll mem[100004][4];

vector < vector < int > > v;



ll solve(int idx , int last) {

    if (idx == n)return 0;



    ll& ret = mem[idx][last];

    if (~ret)return ret;

    ret = 0;

    if (last != 0)ret = max(ret, solve(idx + 1, 0) + v[idx][0]);

    if (last != 1)ret = max(ret, solve(idx + 1, 1) + v[idx][1]);

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

    return ret;

}



int main() {

    //file();



    memset(mem, -1, sizeof mem);

    cin >> n;

    v = vector < vector < int > >(n, vector < int >(3));

    for (auto& e : v)

        for (auto& f : e)

            cin >> f;



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



}

// Faster C++ solution for the same problem (measured speedup: 4.2924467118228735x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

}
// Slower C++ solution:
/******

author :- Mahmoud Tarek.

*******/



#include<bits/stdc++.h>



using namespace std ;



#define rg register

#define dbg(v) cerr << #v << " " << v ;

typedef long long ll ;



int n;

struct doW{

int w[5] ;

};



vector<doW> a ;

int mem[100002][6] ;



int solve(int i, int ch){

if(i == n) return 0 ;

int & ret = mem[i][ch] ;

if(~ret) return ret ;

ret = 0 ;

for(int j = 2 ; j <= 4 ; j++){

    if(ch != j){

        ret = max(ret, a[i].w[j] + solve(i + 1, j)) ;

    }

}

return ret ;

}



int main() {

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

    cin >> n ;

    a.resize(n) ;

    memset(mem, -1, sizeof mem) ;

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

        for(int j = 2 ; j <= 4 ; j++) cin >> a[i].w[j] ;

    }

    cout << solve(0, 1) ;



}



/*

  Empty set of troubles.

*/


// Faster C++ solution for the same problem (measured speedup: 3.448350872162781x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

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

using namespace std;



const int N=1e5+30;

typedef long long ll;

int n;

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

ll dp[N][5];

ll solve(int i,int last){

    if(i==n)return 0;

    int path1=-1 ,path2=-1,path3=-1;

    ll &ret=dp[i][last];

    if(~ret)return ret;

    if(last!=1){

        path1=a[i]+solve(i+1,1);

    }

    if(last!=2){

        path2=b[i]+solve(i+1,2);

    }

    if(last!=3){

        path3=c[i]+solve(i+1,3);

    }

    return ret=max(path1,max(path2,path3));



}

int main()

{

    memset(dp,-1,sizeof dp);

    cin>>n;

    for(int i=0;i<n;i++)cin>>a[i]>>b[i]>>c[i];

    cout<<solve(0,4);



}




// Faster C++ solution for the same problem (measured speedup: 5.145615404143295x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

}
// Slower C++ solution:
/* MaTb3aa */

#include<bits/stdc++.h>

#define ll long long

using namespace std;

int n;

const int N = 100005;

vector<int>v;

ll dp[100005][4];

int A[N], B[N], C[N];

ll solve(int i,int lst) {

	if (i == n)return 0;

	ll& ret = dp[i][lst];

	if (~ret)return ret;

	if (lst != 1)ret = max(ret, solve(i + 1, 1) + A[i]);

	if (lst != 2)ret = max(ret, solve(i + 1, 2) + B[i]);

	if (lst != 3)ret = max(ret, solve(i + 1, 3) + C[i]);

	return ret;

}

int main() {

	cin >> n;

	for (int i = 0; i < n; i++)cin >> A[i] >> B[i] >> C[i];

	memset(dp, -1, sizeof dp);

	cout << solve(0, 0);

	

		 

}

// Faster C++ solution for the same problem (measured speedup: 4.26914856757913x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

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

using namespace std;

int n;

int arr[100005][3];



int dp[100005][4];



int solve(int i, int prev){

	if(i == n) return 0;

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



	dp[i][prev] = -1e7;

	int a = 0, b = 0, c = 0;

	if(prev != 0)

		a = solve(i+1, 0) + arr[i][0];

	if(prev != 1)

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

	if(prev != 2)

		c = solve(i+1, 2) + arr[i][2];



	dp[i][prev] = max(a, max(b, c));

	return dp[i][prev];

}



int main(){

	cin>>n;

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

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



	memset(dp, -1, sizeof dp);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6388972604864644x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

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

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

	cin.tie(NULL), cout.tie(NULL);

}

struct s {

	int x, y, z;

};



int n;

int v[(int)1e5][4];

int dp[(int)1e5][4];



int solve(int i = 0, int pre = 4) {

	if (i >= n)return 0;

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

	int ch1, ch2, ch3;

	ch1 = ch2 = ch3 = 0;

	if (pre != 1) {

		ch1 += solve(i + 1, 1) + v[i][1];

	}

	if (pre != 2) {

		ch2 += solve(i + 1, 2) + v[i][2];

	}

	if (pre != 3) {

		ch3 += solve(i + 1, 3) + v[i][3];

	}

	return dp[i][pre] = max({ ch1,ch2,ch3 });

}





int main()

{

	fast();



	cin >> n;

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

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

	}

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

	cout << solve();



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6507943386226405x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

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

#include<bits/stdc++.h>

using namespace std;



typedef long long ll;



int n;

int a[100005][4];

ll dp[100005][4];



ll solve(int d,int l){

    if(d==n+1) return 0;

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

    ll ret=0;

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

        if(i!=l){

            ret=max(ret, solve(d+1,i)+a[d][i]);

        }

    }

    return dp[d][l]=ret;



}





int main()

{

    cin>>n;

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

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

            cin>>a[i][j];

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

    cout<<solve(1,-1);



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.269242195653212x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

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

#define inf INFINITY

#define file freopen( "input.txt" , "r" , stdin) ; freopen( "output.txt" , "w" , stdout) ;

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

#define _3102_ main()

#define endl '\n'

typedef long long ll;

typedef unsigned long long ull;

typedef long double ld;

using namespace std;

const int N = 1e5 + 100;

const ll MX = 1e9;

const ll mod = 1e9 + 7;

const double eps = 1e-9 ;



ll n, a[N] , b[N] , c[N], dp[N][3] ;



ll solve( int i , int j ){

	if ( i < 0 ) return 0 ;

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

	if ( ret != -1) return ret ;

	if ( j == 0 ){

		return ret = a[i] + max ( solve(i-1 , 1), solve(i-1,2) ) ;

	}

	else if ( j == 1 ){

		return ret = b[i] + max ( solve(i-1 , 0), solve(i-1,2) ) ;

	}

	return ret = c[i] + max ( solve(i-1 , 1), solve(i-1,0) ) ;

}



int _3102_{

	Gamed

	cin >> n ;

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

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

	}

	memset ( dp , -1 , sizeof dp);

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

	cout << max ( solve ( n - 1 , 1) , max ( solve ( n -1 , 0 ) , solve ( n - 1 , 2 ))) << endl ;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.4600806588167474x):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

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



int n, dp1, dp2, dp3;

int tmpa, tmpb, tmpc;



int main() {

   scanf("%d", &n);

   for1(i, n) {

      int a, b, c;

      scanf("%d %d %d", &a, &b, &c);

      if(i == 1) dp1 = a, dp2 = b, dp3 = c;

      else {

         dp1 = max(tmpb, tmpc) + a;

         dp2 = max(tmpa, tmpc) + b;

         dp3 = max(tmpa, tmpb) + c;

      }

      tmpa = dp1, tmpb = dp2, tmpc = dp3;

   }

   printf("%d\n", max(dp1, max(dp2, dp3)));

   return 0;

}