stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.1610194140320624x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    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: 4.129595362434551x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    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: 2.5305069235597824x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    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: 9.214901836119095x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    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: 4.1499601287312595x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    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: 4.126344595589114x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    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: 7.007501533426195x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    cout<<ans<<endl;





    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: 4.10927292865603x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    cout<<ans<<endl;





    return 0;

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

#include <cstring>

#include <stdio.h>

using namespace std;

const int N=1e+5;

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

int memo[N][4];

int dp(int i,int t)

{

    if(i==n)

        return 0;

    int &ret=memo[i][t];

    if(~ret)

        return ret;

    ret=0;

    if(t!=0)

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

    if(t!=1)

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

    if(t!=2)

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

    return ret;





}

int main()

{

    memset(memo,-1,sizeof memo);

    scanf("%d",&n);

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

    {

         scanf("%d %d %d",a+i,b+i,c+i);

    }

    printf("%d\n",dp(0,3));

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.537199345200155x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    cout<<ans<<endl;





    return 0;

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

using namespace std;



#define clr(arr,val) memset(arr,val ,sizeof arr)

//#define first f

//#define second s



typedef  long long ull;

typedef long long ll;

ll const oo=1e18 ;



int const SZ= (int)1e5+5;

//const ll mod=(ll)1e9+7;

    

//int dp[SZ][SZ];

int arr[SZ][4];

int dp[SZ][4];

int n;

int solve(int r ,int prev ){



    if(r>=n){

        return 0;

    }

    int &ret=dp[r][prev];

    if(prev!=-1 && ret !=-1){

        return ret;

    }

    int mx=0;

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

        if(i==prev) continue;

        mx=max(mx,arr[r][i]+solve(r+1,i));

    }

    return ret=mx;

}



int main() {

//#ifdef CLION

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

//    freopen("out.txt", "w", stdin);

//#endif

    

    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,-1)<<endl;

    return 0;

}

 

// Faster C++ solution for the same problem (measured speedup: 2.5477426539833465x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    cout<<ans<<endl;





    return 0;

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

using namespace std;

int main(){

    int n;

    cin>>n;

    long long dp[100005][3];

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

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

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

        long long a,b,c;

        cin>>a>>b>>c;

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

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

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

    }

    long long ans=max(dp[n][0], max(dp[n][1], dp[n][2]));

    cout<<ans<<'\n';

}


// Faster C++ solution for the same problem (measured speedup: 3.356487818160244x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    cout<<ans<<endl;





    return 0;

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



using namespace std;



typedef long long ll;

const double eps=1e-6,pi=acos(-1);

const int N=1e5+5,M=1e9+7,SEGM=4*N,OO=0x3f3f3f3f;

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

int mem[N][5];



int solve(int i=0,int bk=0){

    if(i==n) return 0;

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

    if(~ret) return ret;

    if(bk!=1)

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

    if(bk!=2)

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

    if(bk!=3)

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

    return ret;

}



int main(){

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

    scanf("%d",&n);

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

        scanf("%d%d%d",a+i,b+i,c+i);

    memset(mem,-1,sizeof mem);

    printf("%d",solve());

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.9204306979649153x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    cout<<ans<<endl;





    return 0;

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

#define lld         long long int

#define llf         long double

#define pb          push_back

#define mp          make_pair

#define IN          insert

#define loopN(n)    for(lld i=0;i<n;i++)

#define loop(p,n)   for(lld i=p;i<=n;i++)

#define rloop(p,n)  for(lld i=n;i>=p;i--)

#define loop2(p,n)  for(lld j=p;j<=n;j++)

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

#define F1          first

#define F2          second

#define INF         2e18

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

#define PI          2*acos(0.0)

#define ISO         ios::sync_with_stdio(false);cin.tie(0)

#define mem(a,b)    memset(a,b,sizeof(a))

#define case(z)     cout<<"Case "<<z<<": ";

using namespace std;

const lld mod=1e9+7;



lld on(lld m,lld p){return m | (1<<p);}

bool ison(lld m,lld p){return (bool)(m & (1<<p));}

lld flip(lld m,lld p){return m ^ (1<<p);}

lld allon(lld p){return (1<<p)-1;}

//______________________________________________________________________________________



lld n;

lld dp[100005][3];

lld a[100005][3];



lld fun(lld p, lld k)

{

    if(p>=n)return 0;

    if(dp[p][k]!=-1)

    {

        return dp[p][k];

    }

    lld ans=-INF;

    if(p==0)

    {

        loopN(3)

        {

            ans=max(ans,a[p][i]+fun(p+1,i));

        }

        return ans;

    }

    else

    {

        loopN(3)

        {

            if(i!=k)

            {

                ans=max(ans,a[p][i]+fun(p+1,i));

            }



        }



    }

    return dp[p][k]=ans;







}





int main()

{

    ISO;



    //lld t;cin>>t;while(t--)

    mem(dp,-1);

    cin>>n;

    loopN(n)loop2(0,2)cin>>a[i][j];

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





    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 3.338317169454132x):
#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);





    int n;

    cin>>n;

    vector<int> a(n), b(n), c(n);



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

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

    }



    vector<int> dpA(n), dpB(n), dpC(n);         /// dpA[i] = A was done in i-th day



    dpA[0] = a[0];

    dpB[0] = b[0];

    dpC[0] = c[0];



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

        dpA[i] = a[i] + max(dpB[i-1], dpC[i-1]);

        dpB[i] = b[i] + max(dpA[i-1], dpC[i-1]);

        dpC[i] = c[i] + max(dpA[i-1], dpB[i-1]);

    }





    int ans = max({dpA[n-1], dpB[n-1], dpC[n-1]});



    cout<<ans<<endl;





    return 0;

}