stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 4.238112639987327x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[100005][3];

ll memo[100005][3];



ll solve(int dia, int act){

	if(!dia) return dias[dia][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = 

		dias[dia][act]+max(solve(dia-1, (act+1)%3), solve(dia-1, (act+2)%3));

}



ll solve(){

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

		memo[0][i] = dias[0][i];

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

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

			memo[i][j] = dias[i][j]+

							max(memo[i-1][(j+1)%3], 

								memo[i-1][(j+2)%3]);

	ll maxi = 0;

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

		maxi = max(maxi, memo[N-1][i]);

	return maxi;

}



int main() {

	cin>>N;

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

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

	cout<<solve()<<"\n";

	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: 4.213995471049522x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[100005][3];

ll memo[100005][3];



ll solve(int dia, int act){

	if(!dia) return dias[dia][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = 

		dias[dia][act]+max(solve(dia-1, (act+1)%3), solve(dia-1, (act+2)%3));

}



ll solve(){

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

		memo[0][i] = dias[0][i];

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

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

			memo[i][j] = dias[i][j]+

							max(memo[i-1][(j+1)%3], 

								memo[i-1][(j+2)%3]);

	ll maxi = 0;

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

		maxi = max(maxi, memo[N-1][i]);

	return maxi;

}



int main() {

	cin>>N;

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

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

	cout<<solve()<<"\n";

	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: 7.156353290705878x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[100005][3];

ll memo[100005][3];



ll solve(int dia, int act){

	if(!dia) return dias[dia][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = 

		dias[dia][act]+max(solve(dia-1, (act+1)%3), solve(dia-1, (act+2)%3));

}



ll solve(){

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

		memo[0][i] = dias[0][i];

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

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

			memo[i][j] = dias[i][j]+

							max(memo[i-1][(j+1)%3], 

								memo[i-1][(j+2)%3]);

	ll maxi = 0;

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

		maxi = max(maxi, memo[N-1][i]);

	return maxi;

}



int main() {

	cin>>N;

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

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

	cout<<solve()<<"\n";

	return 0;

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



using namespace std;

typedef long long int lli;



lli n, a[100005], b[100005], c[100005];



lli dp[100005][4];



lli solve(lli i, lli last){

    if(i == n) return dp[i][last] = 0;

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

    lli ans1, ans2, ans3;

    ans1 = ans2 = ans3 = LONG_LONG_MIN;

    if(last != 1) ans1 = a[i] + solve(i+1, 1);

    if(last != 2) ans2 = b[i] + solve(i+1, 2);

    if(last != 3) ans3 = c[i] + solve(i+1, 3);



    return dp[i][last] = max(ans1, max(ans2, ans3));

}





int main(){

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

    int T=1, cs=1;

    //scanf("%d", &T);

    while(T--){

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

        scanf("%lld", &n);

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

            scanf("%lld %lld %lld", &a[i], &b[i], &c[i]);

        }

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



    }



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.196561171641717x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[100005][3];

ll memo[100005][3];



ll solve(int dia, int act){

	if(!dia) return dias[dia][act];

	if(memo[dia][act]) return memo[dia][act];

	return memo[dia][act] = 

		dias[dia][act]+max(solve(dia-1, (act+1)%3), solve(dia-1, (act+2)%3));

}



ll solve(){

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

		memo[0][i] = dias[0][i];

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

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

			memo[i][j] = dias[i][j]+

							max(memo[i-1][(j+1)%3], 

								memo[i-1][(j+2)%3]);

	ll maxi = 0;

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

		maxi = max(maxi, memo[N-1][i]);

	return maxi;

}



int main() {

	cin>>N;

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

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

	cout<<solve()<<"\n";

	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: 3.5242227364639356x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	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: 2.1807911549075336x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	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: 2.260925266631331x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	return 0;

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



using namespace std;



vector<vector<int> > v;

int ans[100005][3];

queue<pair<int,pair<int,int> > >s;

int maior,a=0;



void dp(int i, int bloq, int sum) {

  if(ans[i-1][bloq]>sum)

    return;

  

  if (i>=v.size())

    maior = max(sum, maior);

  else {

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

      if (bloq != j) {



        if(ans[i][j]==-1) {

          ans[i][j] = sum+v[i][j];

          s.push({sum+v[i][j],{i+1,j}});

        }

        else if (ans[i][j]<sum+v[i][j]) {

          ans[i][j] = sum+v[i][j];

          s.push({sum+v[i][j],{i+1,j}});

        }



        //dp(i+1, j, sum+v[i][j]);

      }

    }

  }

}



int main () {

  int n;

  cin>>n;

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

    int a,b,c;

    cin>>a>>b>>c;

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

  }



  maior=0;

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

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

    s.push({v[0][i],{1,i}});

    //dp(1, i, v[0][i]);

  }



  while(!s.empty()) {

    pair<int,pair<int,int> > aux = s.front();

    dp(aux.second.first, aux.second.second, aux.first);

    s.pop();

  }



  cout<<maior<<endl;



}




// Faster C++ solution for the same problem (measured speedup: 2.2106414287345877x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;



int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];



int go(int a, int i){

	if(i == n) return 0;

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

	int& ans = dp[a][i];

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}



int main(){

	cin>>n;

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

	memset(dp, -1, sizeof dp);

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.199892866431997x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	return 0;

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

using namespace std;

const int oo = 1e9;

int number;

const int nx = 1e5+5;

int ac[nx][3];

int dyn[3][nx];

int init(int a, int i){

	if(i == number) 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)+ac[i][0]);

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

	}

	if(a==1){

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

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

	}

	if(a==2){

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

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

	}

	

	return res;

}

int main(){

	cin>>number;

	for(int i=0; i<number; i++) cin>>ac[i][0]>>ac[i][1]>>ac[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: 2.1999168741854396x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;



int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int x, int i){

	if(i==n) return 0;

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

	int& ans = dp[x][i];

	if(x==0)

	{

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(x==1)

	{

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(x==2)

	{

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

	

}

int main(){

    cin>>n;

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

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

    }

    memset(dp, -1, sizeof dp);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.199892866431997x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	return 0;

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

using namespace std;

typedef long long ll;



ll arr[100009][3];

int dp[100009][2][2][2];



long long VAC1(int n,int a, int b, int c)

{



    if(!n && a) return arr[n][0];

    else if(!n && b) return arr[n][1];

    else if(!n && c) return arr[n][2];

    else if(dp[n][a][b][c]!=-1)return dp[n][a][b][c];



    else

    {



        ll x,y,z;

        if(a)

        {

            x = VAC1(n-1,0,1,0)+arr[n][0];

            y = VAC1(n-1,0,0,1)+arr[n][0];



        }



        else if(b)

        {

            x = VAC1(n-1,1,0,0)+arr[n][1];

            y = VAC1(n-1,0,0,1)+arr[n][1];

        }



        else if(c)

        {

            x = VAC1(n-1,1,0,0)+arr[n][2];

            y = VAC1(n-1,0,1,0)+arr[n][2];

        }



        dp[n][a][b][c] = max(x,y);;

        return dp[n][a][b][c];





    }

}



int main()

{

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

    int n;

    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];

        }

    }

    ll res = VAC1(n-1,1,0,0);

    res= max(res, VAC1(n-1,0,1,0));

    res = max(res,VAC1(n-1,0,0,1));



    cout << res << endl;

}


// Faster C++ solution for the same problem (measured speedup: 4.205268421430398x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	return 0;

}
// 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: 2.2125293004093325x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	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: 2.201408915067667x):
#include <bits/stdc++.h>

using namespace std;



#define ll long long int



int N;

int dias[3];

ll memo[2][3];



ll solve(){

	cin>>memo[0][0]>>memo[0][1]>>memo[0][2];

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

		cin>>dias[0]>>dias[1]>>dias[2];

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

			memo[i&1][j]=dias[j]+max(memo[(i+1)&1][(j+1)%3],memo[(i+1)&1][(j+2)%3]);

	}

	ll maxi = 0;

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

		maxi = max(memo[(N-1)&1][j], maxi);

	return maxi;

}



int main() {

	cin>>N;

	cout<<solve()<<"\n";

	return 0;

}