stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <iostream>

#include <cstring>

using namespace std;

const int maxn=1e5+5;

const int inf=0x3f3f3f3f;

int dp[maxn][4];

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

int main()

{

   int n;

   cin>>n;

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

   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]+a[i],dp[i-1][3]+a[i]);

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

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

   }

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

   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.57033699376684x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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;

//const int mod = 1e9+9;

/*ll gcd(ll x, ll y){return (y ? gcd(y, x%y) : x); }

*ll lcm(ll x, ll y){return x*(y/gcd(x,y));}

*void exEuclid(int a, int b){

	if(b==0){x = 1; y=0; d=a; return ;}

	exEuclid(b, a%b);

	int x1=y;

	int y1=x-(a/b)*y;

	x=x1; 100000

	y=y1;

} 

* */

//===========================================================//

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: 2.1777203155474494x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 2.1776983257492177x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 2.552080651476942x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 2.1795198022340934x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 2.5670569786856623x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 2.581372922893813x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 2.1764520646923957x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 2.5762661749107933x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 7.9475980357374425x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 3.3551758493034605x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}
// 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: 2.9759984130305237x):
#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr<<#x<<" = "<<x<<endl;

typedef long long int ll;

int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    int arr[n];

    int arr2[n];

    int arr3[n];

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

        cin>>arr[i];

        cin>>arr2[i];

        cin>>arr3[i];



    }

    int ans[n];

    int ans2[n];

    int ans3[n];

    ans[0]=arr[0];

    ans2[0]=arr2[0];

    ans3[0]=arr3[0];

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

        ans[i]=arr[i]+max(ans2[i-1],ans3[i-1]);

        ans2[i]=arr2[i]+max(ans3[i-1],ans[i-1]);

        ans3[i]=arr3[i]+max(ans2[i-1],ans[i-1]);

    }

    int res=max({ans[n-1],ans2[n-1],ans3[n-1]});

    cout<<res<<"\n";

}