stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.6612765174715527x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 9;

inline int read()

{

    int data = 0, f = 1;

    char ch = getchar();

    while (!isdigit(ch))

    {

        if (ch == '-')

            f = -1;

        ch = getchar();

    }

    while (isdigit(ch))

    {

        data = (data << 3) + (data << 1) + ch - '0';

        ch = getchar();

    }

    return f * data;

}

int a[maxn], b[maxn], c[maxn], dp[maxn][3];

int main()

{

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

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

    //std::ios::sync_with_stdio(false);

    //std::cin.tie(0);

    int n = read();

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

    {

        a[i] = read(), b[i] = read(), c[i] = read();

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

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

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

    }

    printf("%d\n",max(max(dp[n][0], dp[n][1]), dp[n][2]));

    //fclose(stdin);

    //fclose(stdout);

    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: 4.286124290047383x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 9;

inline int read()

{

    int data = 0, f = 1;

    char ch = getchar();

    while (!isdigit(ch))

    {

        if (ch == '-')

            f = -1;

        ch = getchar();

    }

    while (isdigit(ch))

    {

        data = (data << 3) + (data << 1) + ch - '0';

        ch = getchar();

    }

    return f * data;

}

int a[maxn], b[maxn], c[maxn], dp[maxn][3];

int main()

{

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

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

    //std::ios::sync_with_stdio(false);

    //std::cin.tie(0);

    int n = read();

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

    {

        a[i] = read(), b[i] = read(), c[i] = read();

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

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

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

    }

    printf("%d\n",max(max(dp[n][0], dp[n][1]), dp[n][2]));

    //fclose(stdin);

    //fclose(stdout);

    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: 3.4737630421570653x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 9;

inline int read()

{

    int data = 0, f = 1;

    char ch = getchar();

    while (!isdigit(ch))

    {

        if (ch == '-')

            f = -1;

        ch = getchar();

    }

    while (isdigit(ch))

    {

        data = (data << 3) + (data << 1) + ch - '0';

        ch = getchar();

    }

    return f * data;

}

int a[maxn], b[maxn], c[maxn], dp[maxn][3];

int main()

{

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

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

    //std::ios::sync_with_stdio(false);

    //std::cin.tie(0);

    int n = read();

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

    {

        a[i] = read(), b[i] = read(), c[i] = read();

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

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

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

    }

    printf("%d\n",max(max(dp[n][0], dp[n][1]), dp[n][2]));

    //fclose(stdin);

    //fclose(stdout);

    return 0;

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

using namespace std;

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

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

typedef long long ll;

typedef unsigned long long ull;

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

#define forr(i, a, b) for(ll i = (a); i < (ll) (b); i++)

#define forn(i, n) forr(i, 0, n)

#define pb push_back

#define mp make_pair

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

#define DBG(x) cerr << #x << " = " << (x) << endl

#define MOD 1000000007

#define MAXN 100005

ll dp[MAXN][3];

void solve(vector< vector<ll> > &a, int n)

{

    forn(i,MAXN)

    {

        forn(j,3)

        {

            dp[i][j]=(pow(10,9)+5)*(ll)(-1);

        }

    }

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

    forr(i,1,n)

    {

        forn(j,3)

        {

            forn(k,3)

            {

                if(j==k)

                {

                    continue;

                }

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

            }

        }

 

    }

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

}

 

int main() {

    FIN;

    int n;

    cin >> n;

    vector< vector<ll> > arreglo(n);

    forn(i,n) arreglo[i].resize(3);

    forn(i,n)

    {

        forn(j,3) cin >> arreglo[i][j];

    }

    solve(arreglo, n);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.476981521917006x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 9;

inline int read()

{

    int data = 0, f = 1;

    char ch = getchar();

    while (!isdigit(ch))

    {

        if (ch == '-')

            f = -1;

        ch = getchar();

    }

    while (isdigit(ch))

    {

        data = (data << 3) + (data << 1) + ch - '0';

        ch = getchar();

    }

    return f * data;

}

int a[maxn], b[maxn], c[maxn], dp[maxn][3];

int main()

{

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

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

    //std::ios::sync_with_stdio(false);

    //std::cin.tie(0);

    int n = read();

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

    {

        a[i] = read(), b[i] = read(), c[i] = read();

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

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

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

    }

    printf("%d\n",max(max(dp[n][0], dp[n][1]), dp[n][2]));

    //fclose(stdin);

    //fclose(stdout);

    return 0;

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

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

#define ll long long

using namespace std;

const int N = 100010;

int dp[N][3];

int main(){

    fast;

	int n, i, a, b, c;

	cin >> n;

	memset(dp, 0, sizeof dp);

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

		cin >> a >> b >> c;

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

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

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

	}

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.2591206317032557x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 9;

inline int read()

{

    int data = 0, f = 1;

    char ch = getchar();

    while (!isdigit(ch))

    {

        if (ch == '-')

            f = -1;

        ch = getchar();

    }

    while (isdigit(ch))

    {

        data = (data << 3) + (data << 1) + ch - '0';

        ch = getchar();

    }

    return f * data;

}

int a[maxn], b[maxn], c[maxn], dp[maxn][3];

int main()

{

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

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

    //std::ios::sync_with_stdio(false);

    //std::cin.tie(0);

    int n = read();

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

    {

        a[i] = read(), b[i] = read(), c[i] = read();

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

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

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

    }

    printf("%d\n",max(max(dp[n][0], dp[n][1]), dp[n][2]));

    //fclose(stdin);

    //fclose(stdout);

    return 0;

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

using namespace std;



long long w1,w2,hap=0,maxi,aa,bb,cc,N,i,j,k,a[100005][5];

long long M[100005][5];





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cin>>N;

    for(i=0; i<100003; i++)

        for(j=0; j<4; j++)

            M[i][j]=0;



    for(i=0; i<N; i++)

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

    M[N-1][0]=a[N-1][0];

    M[N-1][1]=a[N-1][1];

    M[N-1][2]=a[N-1][2];



    for(i=N-2; i>=0; i--)

    {

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

        {

            if(j==0)

            {

                w1=1;

                w2=2;

            }

            else if(j==1)

            {

                w1=0;

                w2=2;

            }

            else

            {

                w1=0;

                w2=1;

            }

        M[i][j]=a[i][j]+max(M[i+1][w1],M[i+1][w2]);

        }

    }

    aa=max(M[0][0],M[0][1]);

    aa=max(aa,M[0][2]);

    cout<<aa<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.121831065183566x):
#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;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF][3];

long long mem[INF][3];



long long rec(int level,int pre)

{

    if(level==n) return 0;

    if(mem[level][pre]!=-1) return mem[level][pre];

    long long now=0;

    if(pre==0)

    {

        now=max(now,max(a[level][0]+rec(level+1,1),a[level][0]+rec(level+1,2)));

    }else

    if(pre==1)

    {

        now=max(now,max(a[level][1]+rec(level+1,0),a[level][1]+rec(level+1,2)));

    }else

    if(pre==2)

    {

        now=max(now,max(a[level][2]+rec(level+1,0),a[level][2]+rec(level+1,1)));

    }

    return mem[level][pre]=now;

}



int main()

{

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

    scanf("%d",&n);

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

    {

        scanf("%lld%lld%lld",&a[i][0],&a[i][1],&a[i][2]);

    }

    long long ans=max(rec(0,0),max(rec(0,1),rec(0,2)));

    printf("%lld\n",ans);

}


// Faster C++ solution for the same problem (measured speedup: 3.4359660973456974x):
#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<cstdio>

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn = 1e5 + 5;



int a[maxn][4];

int dp[maxn][4];//d[i][j] i day choose j thing 's max happy



int main(){

	int n;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] ,dp[i - 1][3]) + a[i][1];

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

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

	}



	int ans = 0;

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

	printf("%d\n",ans);

}

// Faster C++ solution for the same problem (measured speedup: 1.8115814456617676x):
#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+5;

    int num;

    int act[n][3];

    int dyn[3][n];

    

    int init(int a, int i){

        

    	if(i == num) return 0;

    	

    	if(dyn[a][i] != -1) return dyn[a][i];

    	int& res = dyn[a][i];

    	

    	if(a==0){

    	    

    		res=max(res, init(1, i+1)+act[i][0]);

    		res=max(res, init(2, i+1)+act[i][0]);

    	}

    	

    	if(a==1){

    	    

    		res=max(res, init(0, i+1)+act[i][1]);

    		res=max(res, init(2, i+1)+act[i][1]);

    	}

    	

    	if(a==2){

    	    

    		res=max(res, init(0, i+1)+act[i][2]);

    		res=max(res, init(1, i+1)+act[i][2]);

    	}

    	

    	

    	return res;

    }

    

    int main(){

        

    	cin>>num;

    	for(int i=0; i<num; i++) cin>>act[i][0]>>act[i][1]>>act[i][2];

    	memset(dyn, -1, sizeof dyn);

    	cout<<max(init(0,0),max(init(1,0),init(2,0)))<<"\n";

    	return 0;

    	

    }

// Faster C++ solution for the same problem (measured speedup: 2.236433430514846x):
#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 <stdio.h>

#include <string.h>

#define MAXN 100003



int array_a[MAXN][3];

int array_d[MAXN][3];



int max(int a,int b){

	return a>b? a:b;

}



int main(){

	memset(array_a,-1,sizeof(int)*MAXN*3);

	memset(array_d,-1,sizeof(int)*MAXN*3);

	int N;

	scanf("%d",&N);



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

		scanf("%d %d %d",&array_a[i][0],&array_a[i][1],&array_a[i][2]);

	}



	array_d[0][0]=array_a[0][0];

	array_d[0][1]=array_a[0][1];

	array_d[0][2]=array_a[0][2];

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

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

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

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

	}

	printf("%d",max(array_d[N-1][0],max(array_d[N-1][1],array_d[N-1][2])));

	return 0;



}

// Faster C++ solution for the same problem (measured speedup: 2.624866069200497x):
#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;

}