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

using namespace std;

#define ll long long

ll a[100009],b[100009],c[100009];

int dp[100009][3];

int main() {

    ios::sync_with_stdio(false);

    int n;cin>>n;

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

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

    memset(dp,0x3f, sizeof(dp));

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

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

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

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

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

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.192586659829209x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned ll



#define vi vector <int>

#define ii pair <int, int>

#define vii vector <ii>

#define INFLL 9e18

#define INF 2e9

#define EPS 1e-9

#define pb push_back

#define emp emplace_back

#define PI acos(-1.0)

#define endl "\n"

#define mp make_pair

#define DEBUG 0





int main ()



{

    #define int ll

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    //cout.tie(NULL);



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

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

    

    int n;

    cin >> n;

    int activity[n][3];

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

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

            cin >> activity[i][j];

        }

    }



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

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

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

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

    }



    int ans = 0;

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

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

    }



    cout << ans << endl;

       

    return 0;

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

using namespace std;

const int mx=100010;

int arr[mx][3],n;

int dp[mx][3];

int f(int pos,int ty){

	if(pos==n)return 0;

	int &ans=dp[pos][ty];

	if(ans^-1)return ans;

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

		if(ty^i){

			ans=max(ans,f(pos+1,i)+arr[pos][i]);

		}

	}

	return ans;

}



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

{

	memset(dp,-1,sizeof dp);

	cin>>n;

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

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

			cin>>arr[i][j];

		}

	}

	cout<<max({f(0,0),f(0,1),f(0,2)})<<endl;



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.1815665081196767x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned ll



#define vi vector <int>

#define ii pair <int, int>

#define vii vector <ii>

#define INFLL 9e18

#define INF 2e9

#define EPS 1e-9

#define pb push_back

#define emp emplace_back

#define PI acos(-1.0)

#define endl "\n"

#define mp make_pair

#define DEBUG 0





int main ()



{

    #define int ll

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    //cout.tie(NULL);



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

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

    

    int n;

    cin >> n;

    int activity[n][3];

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

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

            cin >> activity[i][j];

        }

    }



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

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

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

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

    }



    int ans = 0;

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

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

    }



    cout << ans << endl;

       

    return 0;

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

#include<bits/stdc++.h>

#include<unordered_map>

#include<unordered_set>	

using namespace std;

typedef long long ll;

#define bay_bay return 0;

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

#define for1(n) for(int i=1;i<=n;i+=3)

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

#define sz(v) v.size()

#define endl "\n"

constexpr auto EPS = 1e-9;

#define ff fflush(stdout);

const int mod = 1000000007LL;

clock_t T;

#define ctime cerr << "Time : " << double(clock() - T) / CLOCKS_PER_SEC << endl

void GOAT()

{

	std::ios_base::sync_with_stdio(0);

	cin.tie(NULL);

	cout.tie(NULL);

#ifndef ONLINE_JUDGE

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

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

#else

#endif

}

struct day

{

	ll a, b, c;



};

day days[100009];

ll mem[100009][4];

int n;

ll solve(int idx, int ch)

{

	if (idx == n)

	{

		return 0;

	}

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

	if (~ret)

		return ret;

	ret = LLONG_MIN;

	if (ch != 1)

		ret=max(ret, days[idx].a + solve(idx + 1, 1));

	if (ch != 2)

		ret=max(ret, days[idx].b + solve(idx + 1, 2));

	if (ch != 3)

		ret=max(ret, days[idx].c + solve(idx + 1, 3));

	return ret;



}

int main()

{

	//GOAT();

	cin >> n;

	memset(mem, -1,sizeof mem);

	fori(n)

	{

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

	}

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

	

}

// Faster C++ solution for the same problem (measured speedup: 4.168859787318825x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned ll



#define vi vector <int>

#define ii pair <int, int>

#define vii vector <ii>

#define INFLL 9e18

#define INF 2e9

#define EPS 1e-9

#define pb push_back

#define emp emplace_back

#define PI acos(-1.0)

#define endl "\n"

#define mp make_pair

#define DEBUG 0





int main ()



{

    #define int ll

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    //cout.tie(NULL);



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

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

    

    int n;

    cin >> n;

    int activity[n][3];

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

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

            cin >> activity[i][j];

        }

    }



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

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

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

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

    }



    int ans = 0;

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

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

    }



    cout << ans << endl;

       

    return 0;

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

#define p5 100005

using namespace std;



int dp[p5][4],a[p5][4];

int n;



int solve(int day,int state)

{

    if(day==n) return a[day][state];

    if(dp[day][state]!=-1) return dp[day][state];

    int cost=a[day][state];

    if(state==1) cost+=max(solve(day+1,2),solve(day+1,3));

    else if(state==2) cost+=max(solve(day+1,1),solve(day+1,3));

    else if(state==3) cost+=max(solve(day+1,1),solve(day+1,2));

    else cost=max( solve(day+1,1), max(solve(day+1,2),solve(day+1,3)) );



    return dp[day][state]=cost;

}



int main()

{

    scanf("%d",&n);

    for(int i=1;i<=n;i++) scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

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

    printf("%d\n",solve(0,0));



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.554567125661683x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned ll



#define vi vector <int>

#define ii pair <int, int>

#define vii vector <ii>

#define INFLL 9e18

#define INF 2e9

#define EPS 1e-9

#define pb push_back

#define emp emplace_back

#define PI acos(-1.0)

#define endl "\n"

#define mp make_pair

#define DEBUG 0





int main ()



{

    #define int ll

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    //cout.tie(NULL);



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

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

    

    int n;

    cin >> n;

    int activity[n][3];

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

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

            cin >> activity[i][j];

        }

    }



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

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

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

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

    }



    int ans = 0;

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

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

    }



    cout << ans << endl;

       

    return 0;

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

     **author : _KMS_

     **created : 01-04-2020

**/

#include<bits/stdc++.h>

using namespace std;

const long long M =3e5+10,MOD=1e9+7;

typedef long long ll;

vector<int>v[M];

bool vis[M];

int dp[M][7];

int arr[10];

int main()

{

    ios::sync_with_stdio(false);

    cin.tie(0);

    int t,n;

    cin>>n;

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

    {

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

      {

        cin>>arr[j];

       for(int k=1;k<=3;k++)if(j!=k)dp[i][j]=max(dp[i][j],dp[i-1][k]+arr[j]);

      }

    }



  cout<<max(dp[n][1],max(dp[n][2],dp[n][3]))<<"\n";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 9.302517640866354x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned ll



#define vi vector <int>

#define ii pair <int, int>

#define vii vector <ii>

#define INFLL 9e18

#define INF 2e9

#define EPS 1e-9

#define pb push_back

#define emp emplace_back

#define PI acos(-1.0)

#define endl "\n"

#define mp make_pair

#define DEBUG 0





int main ()



{

    #define int ll

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    //cout.tie(NULL);



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

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

    

    int n;

    cin >> n;

    int activity[n][3];

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

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

            cin >> activity[i][j];

        }

    }



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

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

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

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

    }



    int ans = 0;

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

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

    }



    cout << ans << endl;

       

    return 0;

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

#define _CRT_SECURE_NO_WARNINGS

#include <bits/stdc++.h>

#include <unordered_set>

#include <unordered_map>



using namespace std;



typedef long long ll;

typedef long long ull;

typedef pair<ll, ll> ii;



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

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

#define endl		"\n"

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

#define mk			make_pair	



void File()

{

#ifndef ONLINE_JUDGE

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

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

#endif

#ifdef ONLINE_JUDGE

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

	freopen("pyramid.in", "r", stdin);*/

#endif

}

void fast()

{

	ios::sync_with_stdio(NULL);

	cout.tie(NULL);

	cin.tie(NULL);

	//File();

}

const double pi = 2 * acos(0.0);

const ll oo = 0x3f3f3f3f;

const int MOD = 1e9 + 7;

const int nn = 1e5 + 15;

int dx[8] = { 1, -1, 0, 0, 1, -1, 1, -1 };

int dy[8] = { 0, 0, 1, -1, 1, -1, -1, 1 };



int n, k;

vector<pair<int, pair<int, int>>>v;

ll dp[nn][4];

ll solve(int idx, int last)

{

	if (idx >= n)

		return 0;

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

	if (~ret)

		return ret;

	ret = 0;

	if (last != 1)

		ret = max(ret, solve(idx + 1, 1) + v[idx].first);



	if (last != 2)

		ret = max(ret, solve(idx + 1, 2) + v[idx].second.first);



	if (last != 3)

		ret = max(ret, solve(idx + 1, 3) + v[idx].second.second);



	return ret;

}

int main()

{

	fast();

	cin >> n;

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

	{

		int x, y, z;

		cin >> x >> y >> z;

		v.push_back(mk(x, mk(y, z)));

	}

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

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.18941818295845x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned ll



#define vi vector <int>

#define ii pair <int, int>

#define vii vector <ii>

#define INFLL 9e18

#define INF 2e9

#define EPS 1e-9

#define pb push_back

#define emp emplace_back

#define PI acos(-1.0)

#define endl "\n"

#define mp make_pair

#define DEBUG 0





int main ()



{

    #define int ll

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    //cout.tie(NULL);



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

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

    

    int n;

    cin >> n;

    int activity[n][3];

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

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

            cin >> activity[i][j];

        }

    }



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

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

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

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

    }



    int ans = 0;

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

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

    }



    cout << ans << endl;

       

    return 0;

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

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

struct edge

{

    ll  a,b,c;

};

vector<edge> v;

ll  dp[100009][4];

ll  call(ll  ind, ll  pre)

{

    if(ind==v.size()) return 0;

    ll  &ret=dp[ind][pre];

    if(ret!=-1) return ret;

    ret=0;

    ll  a=0,b=0,c=0;

    auto x=v[ind];

        if(pre==1)

        {

            a=x.b+call(ind+1,2);

            b=x.c+call(ind+1,3);

        }

        else if(pre==2)

        {

            a=x.a+call(ind+1,1);

            b=x.c+call(ind+1,3);

        }

        else if(pre==3)

        {

            a=x.a+call(ind+1,1);

            b=x.b+call(ind+1,2);

        }

        else

        {

            a=x.a+call(ind+1,1);

            b=x.b+call(ind+1,2);

            c=x.c+call(ind+1,3);

        }

    return ret=max(a,max(b,c));

}

int  main()

{

    ll  n ;

    cin>>n;

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

    {

        ll  a,b,c; cin>>a>>b>>c;

        v.pb({a,b,c});

    }

    memset(dp,-1,sizeof dp);

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







}


// Faster C++ solution for the same problem (measured speedup: 4.165578111999452x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned ll



#define vi vector <int>

#define ii pair <int, int>

#define vii vector <ii>

#define INFLL 9e18

#define INF 2e9

#define EPS 1e-9

#define pb push_back

#define emp emplace_back

#define PI acos(-1.0)

#define endl "\n"

#define mp make_pair

#define DEBUG 0





int main ()



{

    #define int ll

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    //cout.tie(NULL);



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

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

    

    int n;

    cin >> n;

    int activity[n][3];

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

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

            cin >> activity[i][j];

        }

    }



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

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

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

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

    }



    int ans = 0;

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

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

    }



    cout << ans << endl;

       

    return 0;

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

#define ll long long

using namespace std;



ll n;

std::vector<ll> ara[100010];

ll dp[100010][4];



ll fnc(ll idx,ll last){

  if(idx == n) return 0;

  if(dp[idx][last] != -1) return dp[idx][last];



  ll res1 = - (1<<25);

  ll res2 = - (1<<25);

  ll res3 = - (1<<25);



  if(last == 0){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 1){

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 2){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 3){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

  }

  dp[idx][last] = max(res1, max(res2,res3));

  return dp[idx][last];

}



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

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

  cin >> n;

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

    int a,b,c;

    cin >> a >> b >> c;

    ara[i].push_back(a);

    ara[i].push_back(b);

    ara[i].push_back(c);

  }



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

  return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.074129251988779x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned ll



#define vi vector <int>

#define ii pair <int, int>

#define vii vector <ii>

#define INFLL 9e18

#define INF 2e9

#define EPS 1e-9

#define pb push_back

#define emp emplace_back

#define PI acos(-1.0)

#define endl "\n"

#define mp make_pair

#define DEBUG 0





int main ()



{

    #define int ll

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    //cout.tie(NULL);



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

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

    

    int n;

    cin >> n;

    int activity[n][3];

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

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

            cin >> activity[i][j];

        }

    }



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

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

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

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

    }



    int ans = 0;

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

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

    }



    cout << ans << endl;

       

    return 0;

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

using namespace std;



int n;

int dp[100100][4];

vector<int> arr[100100];



int main() {

    cin >> n;

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

        int k;

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

            cin >> k;

            arr[i].push_back(k);

        }

    }



    for (int i = 0; i < 3; ++i) dp[1][i] = arr[1][i];



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

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



            if ( j == 0 ) 

                dp[pos][j] = arr[pos][j] + max(dp[pos - 1][1], dp[pos - 1][2]);

            

            else if ( j == 1 )

                dp[pos][j] = arr[pos][j] + max(dp[pos - 1][0], dp[pos - 1][2]);



            else

                dp[pos][j] = arr[pos][j] + max(dp[pos - 1][0], dp[pos - 1][1]);

        }

    }



    cout << max({dp[n][0], dp[n][1], dp[n][2]}) << endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.490679014993739x):
/*BISMILLAH

THE WHITE WOLF

NO DREAM IS TOO BIG AND NO DREAMER IS TOO SMALL*/



#include<bits/stdc++.h>

using namespace std;



#define io ios_base::sync_with_stdio(false)

#define ll long long

#define ull unsigned long long

#define vi vector<int>

#define vll vector<long long>

#define pb push_back

#define mod  10000007

#define pii pair<int, int>

#define PI 2*acos(0.0)

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





int main()

{

    io;

    int n;

    cin>>n;

    ll dp[n+1][3];

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

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

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

        dp[0][i] = 0;

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

    {

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

        {

            ll ma = 0;

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

            {

                if(a == j)

                    continue;

                ma = max(ma, dp[i-1][a]);

            }

            dp[i][j] += ma;

        }

    }

    ll ans = 0;

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

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

    cout<<ans;

    return 0;

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

using namespace std;



#define MAX_N 100005

int N, h[MAX_N][3], memo[MAX_N][3];



int main() {

  scanf("%i", &N);

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

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

  	  scanf("%i", &h[d][c]);

	}

  }

  memset(memo, -1, sizeof memo);

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

  	memo[0][c] = h[0][c];

  }

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

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

  	  int &opt = memo[d][c];

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

  	    if (cc != c) {

  	      opt = max(opt, memo[d - 1][cc] + h[d][c]);

		  /*

  	      if (d == 1) {

  	        printf("%i %i %i %i\n", memo[d - 1][cc], h[d][c], cc, c);

		  }

		  */

		}

	  }

	}

  }

  /*

  puts("\ndebug");

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

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

  	  printf("%i ", memo[d][c]);

	}

	puts("");

  }

  puts("end-debug\n");

  */

  int ans = -1;

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

  	ans = max(ans, memo[N - 1][c]);

  }

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

  

}

// Faster C++ solution for the same problem (measured speedup: 2.1600342798303687x):
/*BISMILLAH

THE WHITE WOLF

NO DREAM IS TOO BIG AND NO DREAMER IS TOO SMALL*/



#include<bits/stdc++.h>

using namespace std;



#define io ios_base::sync_with_stdio(false)

#define ll long long

#define ull unsigned long long

#define vi vector<int>

#define vll vector<long long>

#define pb push_back

#define mod  10000007

#define pii pair<int, int>

#define PI 2*acos(0.0)

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





int main()

{

    io;

    int n;

    cin>>n;

    ll dp[n+1][3];

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

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

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

        dp[0][i] = 0;

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

    {

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

        {

            ll ma = 0;

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

            {

                if(a == j)

                    continue;

                ma = max(ma, dp[i-1][a]);

            }

            dp[i][j] += ma;

        }

    }

    ll ans = 0;

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

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

    cout<<ans;

    return 0;

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

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

int main()

{

    int n,i,j,k;

    int a[100005][5];

    scanf("%d",&n);

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

    scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

    int dp[100005][5];

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

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

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

    dp[1][3] = a[1][3];

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

    {

        for(j = 1;j < 4;j++)

        {

            for(k = 1;k < 4;k++)

            {

                if(k != j)

                {

                    if(dp[i][j] < dp[i - 1][k] + a[i][j])

                    dp[i][j] = dp[i - 1][k] + a[i][j];

                }

            }

        }

    }

    printf("%d\n",max(dp[n][1],max(dp[n][2],dp[n][3])));

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2394056712254784x):
/*BISMILLAH

THE WHITE WOLF

NO DREAM IS TOO BIG AND NO DREAMER IS TOO SMALL*/



#include<bits/stdc++.h>

using namespace std;



#define io ios_base::sync_with_stdio(false)

#define ll long long

#define ull unsigned long long

#define vi vector<int>

#define vll vector<long long>

#define pb push_back

#define mod  10000007

#define pii pair<int, int>

#define PI 2*acos(0.0)

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





int main()

{

    io;

    int n;

    cin>>n;

    ll dp[n+1][3];

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

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

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

        dp[0][i] = 0;

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

    {

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

        {

            ll ma = 0;

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

            {

                if(a == j)

                    continue;

                ma = max(ma, dp[i-1][a]);

            }

            dp[i][j] += ma;

        }

    }

    ll ans = 0;

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

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

    cout<<ans;

    return 0;

}