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

#define ll long long

using namespace std;

const int maxn = 1e5+5;

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



int main(){

	int n; cin >> n;

	for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];

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

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

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

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

	}

	int ans;

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

	cout << ans << endl;

}

// Faster C++ solution for the same problem (measured speedup: 4.496442425668509x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  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: 18.964984417104166x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  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: 11.621241347092411x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

using namespace std;

const int maxn = 1e5+10;

int dp[maxn][4];

int main()

{

    int n;

    cin>>n;

    int ans =0;

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

    {

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

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

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

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

        ans = max(dp[i][1],max(dp[i][2],dp[i][3]));

    }

    cout<<ans<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 4.495717226178184x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  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: 42.31902992648543x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include <bits/stdc++.h>

#define mod 998244353

typedef long long ll;



using namespace std;



void Fastio(){

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

    if (exp == 0)

        return 1;

    ll ans = fp(base, exp / 2);

    ans = (ans * ans) % mod;

    if (exp % 2 != 0)

        ans = (ans * (base % mod)) % mod;

    return ans;

}



void calcFacAndInv(ll n)

{

    fact[0] = inv[0] = 1;

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

    {

        fact[i] = (i * fact[i - 1]) ;

        inv[i] = fp(fact[i], mod - 2);

    }

}



ll ncr(ll n, ll r)

{

    return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

    return (fact[n] * inv[n - r]) % mod;

}



long double  NCR(int n, int m){

    long double res = 1;

    m = min(m, n - m);

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

        res = res * ( n - i) / (i + 1);

 }

    return res;

}

ll tol(string a){

    ll res=0;

    while(a[0]=='0'){

        a.erase(a.begin());

    }

    if(a=="")a="0";

    for(int i=0; i<a.size(); i++){

        res*=10; res+=a[i]-'0';

    }

    return res;

}

int check(string a, string b){

    for(int i=0; i<a.size(); i++){

        if(a[i]>b[i])return 1;

        if(b[i]>a[i])return -1;

    }

    return 0;

}

class data{

public:

    ll v,d,p;

    data(){}

    data(ll vv, ll dd, ll pp){

        v=vv;  d=dd; p=pp;

    }

};



int primeFactors(int n)

{

    ll c=0;

    while (n % 2 == 0)

    {

        c++;

        n = n/2;

    }

    for (int i = 3; i*i <= n; i = i + 2)

    {

        while (n % i == 0)

        {

            c++;

            n = n/i;

        }

    }

    if (n > 2)

        c++;

    return c;

}



int main()

{

    Fastio();

    int n; cin>>n;

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

    ll dp[n+1][5];

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

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

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

    }

    for(int i=n-1; i>=0; i--){

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

            ll res1,res2,res0;

            res0=res1=res2=-1e9;

            res0=dp[i+1][0]+a[i];

            res1=dp[i+1][1]+b[i];

            res2=dp[i+1][2]+c[i];

            if(y==0){

                dp[i][y]=max(res1,res2);

            }

            if(y==1){

                dp[i][y]=max(res0,res2);

            }

            if(y==2){

                dp[i][y]=max(res0,res1);

            }

        }

    }

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





    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.556562949338748x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

using namespace std;

const int N = 2e5 ;

int dpA[N], dpB[N] , dpC[N] ,a[N] , b[N] , c[N],n;

int main()

{

    cin >> n ;

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

    dpA[0] = a[0] ;

    dpB[0] = b[0] ;

    dpC[0] = c[0] ;

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

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

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

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

    }

    cout << max(dpA[n-1],max(dpB[n-1],dpC[n-1]));

}


// Faster C++ solution for the same problem (measured speedup: 4.489827727662909x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include <vector>

#include <string>

#include <cmath>

#include <set>

#include <stdio.h>

#include <algorithm>

#include <map>

#include <queue>

#include <unordered_map>

#define ll long long

#define pii pair <int, int>

#define pll pair <long long, long long>

#define plxy pair <long long, pll>

#define INT_MAX 1e18

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

using namespace std;

using namespace std;

const int mod=1e9+7;

const int N=1e5+5;

ll ans[N][3];

int main()

{

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

     cin>>n;

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

     {

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

     }

     ans[0][0]=a[0];

     ans[0][1]=b[0];

     ans[0][2]=c[0];

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

     {

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

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

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

     }

     cout<<max(ans[n-1][0],max(ans[n-1][1],ans[n-1][2]));

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.68851874431934x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#define ll long long

using namespace std;



int main()

{

    int n;

    cin>>n;

    int a[n+5],b[n+5],c[n+5];



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

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

    int dp[3][n+5];

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

    dp[1][1]=b[1];

    dp[2][1]=c[1];

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

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

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

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

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.455475513521625x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include <bitset>

using namespace std;

#define ll long long

#define mod 1000000009



int arr[100005][3];

ll sol[100005][3];



int main()

{

    ios_base::sync_with_stdio(false);    cin.tie(NULL);



    ll n; cin >> n;

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

        for(ll ii=0; ii<3; ii++){

            cin >> arr[i][ii];

        }

    }

    sol[0][0] = arr[0][0];

    sol[0][1] = arr[0][1];

    sol[0][2] = arr[0][2];



    ll maxi = max(arr[0][0],max(arr[0][1],arr[0][2]));

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

        for(ll ii=0; ii<3; ii++){

            maxi = 0;

            for(ll x=0; x<3; x++){

                if(x == ii) continue;



                maxi = max(maxi,arr[i][ii] + sol[i-1][x]);

            }

            sol[i][ii] = maxi;

        }

    }









    cout << max(sol[n-1][0],max(sol[n-1][1],sol[n-1][2]));







    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.554934004030819x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

}
// Slower C++ solution:
#define f(i,s,v) for(int i=(int)s;i<(int)v;++i)

#define rf(i,s,v) for(int i=(int)s-1;i>(int)v;--i)

#include<bits/stdc++.h>

typedef long long ll;

using namespace std;

const ll oo=1e15;

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

char A='a',B='b',C='c';

map<pair<int,char>,ll>calc;

ll vacation(int day,char last){

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

    if(calc[{day,last}])

        return calc[{day,last}];

    ll maxi=0;

    if(last!=A)

        maxi=max(maxi,a[day]+vacation(day+1,A));

    if(last!=B)

        maxi=max(maxi,b[day]+vacation(day+1,B));

    if(last!=C)

        maxi=max(maxi,c[day]+vacation(day+1,C));

    calc[{day,last}]=maxi;

    return maxi;

}

int main()

{

    scanf("%d",&n);

    f(i,1,n+1)

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

    cout<<vacation(1,'Z')<<'\n';

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.512734520662895x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

}
// Slower C++ solution:
//Author: Fuadul Hasan(fuadul202@gmail.com)

//BSMRSTU,Gopalganj

#include<bits/stdc++.h>

using namespace std;int tc = 1;

#define happy ios::sync_with_stdio(false);

#define coding  cin.tie(0);

#define mp make_pair

#define ll long long

#define Pi atan(1)*4

#define pb  push_back

#define vpr vector<pr> 

#define pr pair<int, int> 

#define vi std::vector<int>

#define vll std::vector<ll>

#define YES printf("YES\n");

#define NO printf("NO\n");

#define Yes printf("Yes\n");

#define No printf("No\n");

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

#define point(x) setprecision(x)

#define Test printf("Case %d: ",tc++);

#define Unique(c) (c).resize(unique(all(c))-(c).begin())



const int M = 1e9 + 7;

const ll  Inf = (ll)2e18 + 5;

const int N   = 2e5 + 5;

ll dp1[N],dp2[N],dp3[N];

int solve() 

{

	//happy coding



	int n;

	cin>>n;

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

		int a,b,c;

		cin>>a>>b>>c;

		dp1[i] = max(dp2[i-1]+a,dp3[i-1]+a);

		dp2[i] = max(dp1[i-1]+b,dp3[i-1]+b);

		dp3[i] = max(dp1[i-1]+c,dp2[i-1]+c);

	}

	int x = max(dp1[n],max(dp2[n],dp3[n]));

	cout<<x<<endl;

	return 0;

}

int main(){

	int test = 1;

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

	while (test--)solve();return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.493365239564085x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  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: 19.058508707398637x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include<cmath>

#include<iostream>

#define maxn 100005

using namespace std;



int main(void)

{

    int N;

    cin >> N;

    int a[maxn][3];

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

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

    }

    int dp[N+1][3];

    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++){

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

        }

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.593727203017022x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  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: 18.950055413858912x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

}