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

using namespace std;

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: 2.207326509360405x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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

using namespace std;



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



int n;

int arr[3][N];



int mem[N][4];

int solve(int i, int last)

{

    if(i==n)

        return 0;

    if(mem[i][last]!=-1)

        return mem[i][last];

    int op1= (last!=1) ? solve(i+1, 1)+arr[0][i] : 0;

    int op2= (last!=2) ? solve(i+1, 2)+arr[1][i] : 0;

    int op3= (last!=3) ? solve(i+1, 3)+arr[2][i] : 0;

    return mem[i][last]=max(op1,max(op2,op3));

}





int main()

{

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

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

    scanf("%d",&n);

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

        scanf("%d %d %d",arr[0]+i,arr[1]+i,arr[2]+i);

    memset(mem, -1, sizeof mem);

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







    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.586802409416716x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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

using namespace std;

#define ALL(x) begin(x),end(x)

typedef long long ll;

int OO = 0x3f3f3f3f;



int n;

int a[100005];

int b[100005];

int c[100005];

int mem[100005][3];



int solve(int x, int k)

{

    int op1, op2, op3;

    if(x == n)

        return 0;



    if(mem[x][k] != -1)

        return mem[x][k];



    op1 = op2 = op3 = 0;

    if(k != 1)

        op1 = solve(x+1, 1)+a[x];

    if(k != 2)

        op2 = solve(x+1, 2)+b[x];

    if(k != 3)

        op3 = solve(x+1, 3)+c[x];



    return mem[x][k] = max(op1, max(op2, op3));

}



int main()

{

    memset(mem, -1, sizeof mem);

    cin >> n;

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

    {

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

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.2091727675329453x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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



using namespace std;



const int N = 1e5 + 9, OO = 0x3f3f3f3f;



int n, points[N][3], dp[N][4];



int solve(int i, int prev){

    if(i == n)  return 0;



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

    if(~ret)    return ret;



    ret = 0;

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

        if(prev != j)   ret = max(ret, points[i][j] + solve(i + 1, j));





    return ret;

}



int main(){



    cin >> n;

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

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



    memset(dp, -1, sizeof dp);



    cout << solve(0, 3);

}


// Faster C++ solution for the same problem (measured speedup: 2.601982493668878x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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

#include <algorithm>

#include <iostream>

#include <iomanip>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <vector>

#include <string>

#include <climits>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <deque>

#include <stdio.h>

//#define int long long

using namespace std;

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

#define endl '\n'

#define OnlineJudgec

#define F first

#define S second

#define  precision(n)  cout << fixed << setprecision(n)

#define clr(v,d ) memset(v, d, sizeof(v))

typedef long long ll;

typedef vector<int> vi;

void init()

{

    cin.tie(0);

    cin.sync_with_stdio(0);

    cout.tie(0);

    #ifdef OnlineJudge

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

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

    #endif

}

ll S(ll x){ll s = 0;while(x){s += x % 10;x /= 10;}return s;} /// sum of digits.

ll powe(ll a , ll n){if(n == 1)return a;return a * powe(a ,n-1);}

ll com(int n,int m){ll z = 1;for(int i = 0; i < m;i++){z = z * ( n - i) / (i + 1) ;}return z;}

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

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

int dr[] = {0,0,0,1,1,1,-1,-1,-1};

int dc[] ={-1,0,1,-1,0,1,-1,0,1};

/// <<------------------------------------------------------------------------------>>

const int mx = 100009;

int n , k;

int arr[mx][4];

int dp[mx][4];

int solve(int idx , int last)

{

     if(idx == n)return 0;

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

     if(~ret) return ret;

     int A = 0 ,B = 0 ,C = 0;

     if(last != 0)

        A = solve(idx+1 , 0) + arr[idx][0];

     if(last != 1)

        B = solve(idx+1 , 1) + arr[idx][1];

     if(last != 2)

        C = solve(idx+1 , 2) + arr[idx][2];

     return ret = max(A,max(B,C));

}

/// <<------------------------------------------------------------------------------>>

int main()

{

    init();

    cin >> n;

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

    {

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

    }

    clr(dp,-1);

    cout<<solve(0,3);

}


// Faster C++ solution for the same problem (measured speedup: 2.616493210228438x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

}
// Slower C++ solution:
/*

██╗  ██╗ █████╗ ██████╗ ███████╗███████╗███╗   ███╗        ██╗  ██╗██╗  ██╗ █████╗ ██╗     ███████╗██████╗

██║ ██╔╝██╔══██╗██╔══██╗██╔════╝██╔════╝████╗ ████║        ██║ ██╔╝██║  ██║██╔══██╗██║     ██╔════╝██╔══██╗

█████╔╝ ███████║██████╔╝█████╗  █████╗  ██╔████╔██║        █████╔╝ ███████║███████║██║     █████╗  ██║  ██║

██╔═██╗ ██╔══██║██╔══██╗██╔══╝  ██╔══╝  ██║╚██╔╝██║        ██╔═██╗ ██╔══██║██╔══██║██║     ██╔══╝  ██║  ██║

██║  ██╗██║  ██║██║  ██║███████╗███████╗██║ ╚═╝ ██║███████╗██║  ██╗██║  ██║██║  ██║███████╗███████╗██████╔╝

╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═╝     ╚═╝╚══════╝╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═════╝

Believe in yourself ,( try and try and then try ), You can do it 

*/

#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<unordered_map>

#define endl "\n"

#define Ceil(x,y) ((x+y-1)/y)

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

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

#define allr(s) s.rbegin(),s.rend()

#define rt(s) return cout<<s,0

#define watch(x) cout<<(#x)<<" = "<<x<<endl

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

#define OO 0x3f3f3f3f3f3f3f3fLL

const int oo = 0x3f3f3f3f, mod = 1e9 + 7;

using namespace std;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

//x = xDegrees*3.14159265 / 180;

//ef1tn=(( (n / 2)*( (n / 2) + 1) / 2) * 2);

//of1tn=(n * (n + 1) / 2 - 2 * ((n / 2) * (n / 2 + 1) / 2));

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

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

void K_K(){

#ifndef ONLINE_JUDGE

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

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

#endif  !ONLINEJUDGE

	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

}

ll gcd(ll a, ll b) {

	return b == 0 ? abs(a) : gcd(b, a % b);

}

ll lcm(ll a, ll b) {

	return a / gcd(a, b) * b;

}

int n;

vector<int>a, b, c;

int dp[100005][5];

int sol(int idx, int pre){

	if (idx == n)

		return 0;

	int& ret = dp[idx][pre];

	if (~ret)

		return ret;

	int ch1 = 0, ch2 = 0, ch3 = 0;

	if (pre != 1)

		ch1 = sol(idx + 1, 1) + a[idx];

	if (pre != 2)

		ch2 = sol(idx + 1, 2) + b[idx];

	if (pre != 3)

		ch3 = sol(idx + 1, 3) + c[idx];

	return ret = max({ ch1, ch2, ch3 });

}

int main(){

	K_K();

	cin >> n;

	a.resize(n);

	b.resize(n);

	c.resize(n);

	clr(dp, -1);

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

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

	cout << sol(0, 0);

}

// Faster C++ solution for the same problem (measured speedup: 3.0186369569328324x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define     watch(a)        cout<<fixed<<setprecision(a)

#define     lp(i,a,b)       for(int i=a;i<b;i++)

#define     Endl            "\n";

#define     tab             "\t"

#define     reset(a,b)      memset(a,b,sizeof(a));

#define     sf(a)           scanf(" %d", &a);

#define     srt(a)          sort(a.begin(),a.end());

#define     pi              2*acos(0.0)

#define     inf             INT_MAX

const int sz=1e5+9;

int n,x,arr[sz][3],dp[sz][3];



int clac(int i,int ignore ,int value){

    if(i>n)return value;

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

    int ret=0;

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

        if(j==ignore)

            continue;

        ret=max(ret,value+clac(i+1,j,arr[i][j]));

    }

    return dp[i][ignore]=ret;

}



void solve(){

    cin>>n;

    reset(dp,-1);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    lp(i,1,n+1)cin>>arr[i][0]>>arr[i][1]>>arr[i][2];

    int ans1=clac(2,0,arr[1][0]);

    //cout<<dp[2][0]<<tab<<dp[2][1]<<tab<<dp[2][2]<<endl;

    int ans2=clac(2,1,arr[1][1]);

    int ans3=clac(2,2,arr[1][2]);

    //cout<<ans1<<tab<<ans2<<tab<<ans3<<endl;

    //int ans=max(ans1,(max(ans2,ans3)));

   // bug(ans1,1);

    cout<<max(ans1,(max(ans2,ans3)))<<"\n";

}



int main(){

    //FasterIO;

    #ifndef ONLINE_JUDGE

       //freopen("1.txt","r",stdin);

    #endif

    // int t;

    //cin>>t;

    //while(t--)

        solve();

}


// Faster C++ solution for the same problem (measured speedup: 2.2060632925797425x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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

using namespace std;



#define ll long long

#define size 100001



int a[size];

int b[size];

int c[size];

int dp[size][4];



int max_sum(int n, int p)

{

    if(n <= 0)

        return 0;

    if(n > 0 && dp[n][p] != -1)

        return dp[n][p];

    int ans = INT_MIN;

    if(p != 0)

    {

        ans = max(ans,a[n]+max_sum(n-1,0));

    }

    if(p != 1)

    {

        ans = max(ans,b[n]+max_sum(n-1,1));

    }

    if(p != 2)

    {

        ans = max(ans,c[n]+max_sum(n-1,2));

    }



    return dp[n][p] = ans;

}







int main()

{

  //OJ;

  int t = 1;

  //cin>>t;

  while(t--){

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

  int n;

  cin>>n;



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

  {

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

  }

  cout<<max_sum(n,3)<<endl;

  }

   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.6113169835370464x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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





using namespace std;



vector<vector<int>> vec(100007);

int n;



long long meme[100007][5];





long long solve(int index, int flag){

    if(index==n){

        return 0;

    }

    

    long long &res = meme[index][flag]; 

    

    if(res != -1){

        return res;

    }

    res = 0;

    

    if(flag!=0){

        res = max(res,vec[index][0]+solve(index+1,0));

    }

    if(flag!=1){

        res =  max(res,vec[index][1]+solve(index+1,1));

    }

    if(flag!=2){

        res =  max(res,vec[index][2]+solve(index+1,2));

    }

    return res;

}



int main(){

    cin>>n;

    int num1,num2,num3;

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

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

        cin>>num1>>num2>>num3;

        vec[i].push_back(num1);

        vec[i].push_back(num2);

        vec[i].push_back(num3);

    }

    long long sol = solve(0,3);

    cout<<sol<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 8.055727289035062x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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

#define inf 1000000007

#define N 100007

using namespace std;



int a[N];

int b[N];

int c[N];

int n;

long long mem[N][3];



long long dp(int i, int j)

{

    if(i >= n)

        return 0;

    if(mem[i][j] != -1)

        return mem[i][j];

    long long ans = 0;

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

    {

        if(k == 0 && k != j)

        {

            long long val = a[i] + dp(i + 1, k);

            ans = max(ans, val);

        }

        if(k == 1 && k != j)

        {

            long long val = b[i] + dp(i + 1, k);

            ans = max(ans, val);

        }

        if(k == 2 && k != j)

        {

            long long val = c[i] + dp(i + 1, k);

            ans = max(ans, val);

        }

    }

    mem[i][j] = ans;

    return ans;

}



int main()

{

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

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

    cin >> n;

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

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

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

    long long val1 = dp(0, 0);

    long long val2 = dp(0, 1);

    long long val3 = dp(0, 2);

    long long ans = max(val1, val2);

    ans = max(ans, val3);

    cout << ans << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.4008239379002974x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



int f(int idx, int now) {

    if(idx == n) return 0;

    int &sol = dp[idx][now];

    if(sol != -1) return sol;

    if(now == 0) {

        sol = max(f(idx + 1, 1) + a[idx + 1], 

              max(f(idx + 1, 2) + b[idx + 1], 

                  f(idx + 1, 3) + c[idx + 1]));

    } else if(now == 1) {

        sol = max(f(idx + 1, 2) + b[idx + 1],

                  f(idx + 1, 3) + c[idx + 1]);

    } else if(now == 2) {

        sol = max(f(idx + 1, 1) + a[idx + 1],

                  f(idx + 1, 3) + c[idx + 1]);

    } else {

        sol = max(f(idx + 1, 1) + a[idx + 1],

                  f(idx + 1, 2) + b[idx + 1]);

    }

    return sol;

}



void solve() {

    cin >> n;

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

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

    }

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

    cout << f(0, 0) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}

// Faster C++ solution for the same problem (measured speedup: 3.0164876885033025x):
#include<bits/stdc++.h>

using namespace std;

#define  ll long long int

#define mx 100005

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

ll dp[mx][4];

int main()

{

    ll n,k;

    cin>>n;



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

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

    }

    dp[1][1]=a[1];dp[1][2]=b[1];dp[1][3]=c[1];

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

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

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

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

    }

    cout<<max({dp[n][1],dp[n][2],dp[n][3]})<<endl;

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

using namespace std;



int ar[100010][5],n;

int dp[100010][5];



int call(int i,int f)

{



    if(i==n)

        return 0;



    if(dp[i][f]!=-1)

        return dp[i][f];



    int ans=0;



    if(f==0)

        ans+=max(ar[i][1]+call(i+1,1),ar[i][2]+call(i+1,2));



    else if(f==1)

        ans+=max(ar[i][0]+call(i+1,0),ar[i][2]+call(i+1,2));

    else

        ans+=max(ar[i][1]+call(i+1,1),ar[i][0]+call(i+1,0));



        dp[i][f]=ans;

        return ans;



}



int main()

{



    cin>>n;



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

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



    memset(dp,-1,sizeof dp);



    int ans=max(max(call(0,0),call(0,1)),call(0,2));



    cout<<ans<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 2.9409663777777526x):
#include<bits/stdc++.h>

using namespace std;



#define ff              first

#define ss              second

#define ll             long long

#define pb              push_back

#define mp              make_pair

#define pii             pair<ll,ll>

#define vi              vector<ll>

#define mii             map<ll,ll>

#define pqb             priority_queue<ll>

#define pqs             priority_queue<ll,vi,greater<ll> >

#define setbits(x)      __builtin_popcountll(x)

#define zrobits(x)      __builtin_ctzll(x)

#define mod             1000000007

#define inf             1e18

#define ps(x,y)         fixed<<setprecision(y)<<x

#define mk(arr,n,type)  type *arr=new type[n];

#define w(x)            ll x; cin>>x; while(x--)

#define FIO             ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)



struct activity{

ll A,B,C;};



ll solve(ll n,vector<activity>v){

    ll dp[n+1][3];

dp[1][0]=v[1].A;

dp[1][1]=v[1].B;

dp[1][2]=v[1].C;

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

{

    dp[i][0]=v[i].A+max(dp[i-1][1],dp[i-1][2]);

    dp[i][1]=v[i].B+max(dp[i-1][0],dp[i-1][2]);

    dp[i][2]=v[i].C+max(dp[i-1][1],dp[i-1][0]);

}



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



}



int main()

{

    FIO;

    ll a,b,c,d,e,f=0,sum=0,ans,ck=0,m,n,k,p,q,r,t,add=0,x,y,z;



    string s;



    cin>>n;

    vector<activity>v(n+1);

    for(int i=1;i<=n;i++){cin>>v[i].A;cin>>v[i].B;cin>>v[i].C;}

    ans=solve(n,v);

    cout<<ans<<endl;







    return 0;

}