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



using namespace std;

typedef long long ll;

typedef double dd;

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

#define endl "\n"

#define clr(n, r) memset(n,r,sizeof(n));



void fast() {

    cin.tie(0);

    cin.sync_with_stdio(0);

}

const int MAX=1e5+10;

int n,a[MAX],b[MAX],c[MAX],dp[MAX][3];

int ans(int i,int prev){

    if(i==n)return 0;

    int path1=0,path2=0,path3=0;

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

    if(prev!=-1&&ret!=-1)return ret;

    if(prev!=0){

        path1=a[i]+ans(i+1,0);

    }

    if(prev!=1){

        path2=b[i]+ans(i+1,1);

    }

    if(prev!=2){

        path3=c[i]+ans(i+1,2);

    }

    return ret=max(path1,max(path2,path3));

}



int main() {

    fast();

    cin>>n;

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

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

    }

    clr(dp,-1);

    cout<<ans(0,-1);

}


// Faster C++ solution for the same problem (measured speedup: 2.186038477794119x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

}
// Slower C++ solution:
/*

** author:mahmoud magdy

*/

#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define endl "\n"



inline int D()

{

    int m ;

    cin>>m;

    return m ;

}

inline ll lD()

{

    ll m ;

    cin>>m;

    return m ;

}

inline double dD()

{

    double m;

    cin>>m;

    return m;

}

void fast()

{

    ios_base :: sync_with_stdio( 0 ) ;

    cin.tie( 0 ) ;

    cout.tie( 0 ) ;

}

int n;

ll a[100005],b[100005],c[100005];

ll dp[100005][5];

ll solve(int i,int prv){

 if(i==n)return 0;

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

 ll x=0;

 if(prv!=1){

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

 }

 ll y=0;

 if(prv!=2)

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

 ll z=0;

 if(prv!=3)

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

 return dp[i][prv]=max({x,y,z});

}

int main()

{

    memset(dp,-1,sizeof dp);

    fast();

    n=D();

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

        a[i]=D();b[i]=D();c[i]=D();

    }

    cout<<solve(0,10)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.028418884806515x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

}
// Slower C++ solution:


#include <bits/stdc++.h>

using namespace  std;



int N;

const int SIZE = 100005;

int mem [SIZE][4];

int arr [SIZE][3];





int dp (int i, int activ) {

  

  if (i == N) return 0;

  



 if (mem[i][activ] != -1) return mem[i][activ];  

  

  int ret = -1; 



  if (activ != 0) {

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

  }



  if (activ != 1) {

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

    

  }



  if (activ != 2) {

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

    

  }



  return  mem[i][activ] = ret;

}



int main (){ 



  cin  >> N ;

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

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

  }



  memset(mem, -1, sizeof mem);

  





  cout << dp(0,3);



}

// Faster C++ solution for the same problem (measured speedup: 2.565002111276256x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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



using namespace std;



int a[100007], b[100007], c[100007];

int memo[100007][4];

int n; 



int solve(int p, int last) {

    if(p == n) return 0;

    int & st = memo[p][last];

    if(st == -1) {

        st = max({

            last != 1? solve(p+1, 1) + a[p] : 0,

            last != 2? solve(p+1, 2) + b[p] : 0,

            last != 3? solve(p+1, 3) + c[p] : 0

        });

    }

    return st;

}

int main() {

    memset(memo, -1, sizeof memo);

    cin >> n;

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

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

    }



    cout << solve(0, 0);

}

// Faster C++ solution for the same problem (measured speedup: 2.57129515888047x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

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

int dp[MXN][4];



int Dp(int d, int lst) {

    if(d == n) return 0;

    if(dp[d][lst] != -1) return dp[d][lst];

    int ans = 0;

    if(lst != 0) ans = Dp(d + 1, 0) + a[d];

    if(lst != 1) ans = max(ans, Dp(d + 1, 1) + b[d]);

    if(lst != 2) ans = max(ans, Dp(d + 1, 2) + c[d]);

    return dp[d][lst] = ans;

}



int main() {

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

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

    cout << Dp(0, 3) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.5793153855105437x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

    if(dp[y][x]!=-1)

        return dp[y][x];

    if(y == 0 )

    {

    	dp[y][x] = A[y][x];

        return A[y][x];

    }

    return dp[y][x] = max(solver((y-1),(x+1)%3)+ A[y][x],

	solver((y-1),(x+2)%3) + A[y][x]);

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.1752817417333326x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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

 

#define pb push_back

#define fst first

#define snd second

#define INF 0x7fffffff

#define INFLL 0x7FFFFFFFFFFFFFFF

#define lenA(ar) sizeof(ar)/sizeof(*ar)

#define lenrows(mat) sizeof(mat)/sizeof(*mat)

#define lencols(mat) sizeof(*mat)/sizeof(**mat)

#define fill(a,c) memset(&a, c, sizeof(a))

#define fore(i,x,y) for(ll i=x;i<y;i++)

#define foreI(i,x,y) for(ll i=x-1;i>=y;i--)

#define max2(a,b) max(a, b)

#define max3(a,b,c) max2(max2(a,b),c)

#define min2(a,b) min(a, b)

#define min3(a,b,c) min2(min2(a,b),c)

#define MOD 1000000009

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

#define rall(x) (x).rbegin(), (x).rend()

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

 

typedef long long ll;

typedef unsigned long long ull;

 

using namespace std;

ll arr[100010][3];

ll dp[100010][3];

int n;

ll solve(int i, int j){

    if(i==n-1) return arr[i][j];

    ll &r = dp[i][j];

    if(r>-1) return r;

    ll a = 0; ll b = 0;ll c = 0; ll act=arr[i][j];

    if(j!=0) a=solve(i+1, 0)+act;

    if(j!=1) b=solve(i+1, 1)+act;

    if(j!=2) c=solve(i+1, 2)+act;

   return r=max3(a,b,c);

}

 

int main() {

     

    cin>>n;

    arr[n][3];

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

    fore(i,0,n){

        fore(j,0,3){

            cin>>arr[i][j];

        }

    }

 

    cout<<max3(solve(0,0),solve(0,1),solve(0,2))<<"\n";

 

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.3576548562997446x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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

#include <iostream>

using namespace std;

const int MAXN=1e5+7;

int n,ans;

int h[MAXN][5],f[MAXN][5][5];

int main()

{

	cin>>n;

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

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

	memset(f,127,sizeof(f));

	for(int i=1;i<=3;i++)

		for(int j=1;j<=3;j++) f[1][i][j]=h[1][i];

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

		for(int j=1;j<=3;j++)

			for(int k=1;k<=3;k++)

			{

				int maxx=0;

				for(int l=1;l<=3;l++)

					if(j!=k&&k!=l) maxx=max(maxx,f[i-1][k][l]);

				f[i][j][k]=maxx+h[i][j];

			}

	for(int i=1;i<=3;i++)

		for(int j=1;j<=3;j++) ans=max(ans,f[n][i][j]);

	cout<<ans<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 11.361250186511171x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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

using namespace std;

#define sz 100005

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

int dp[sz][5];



int res(int ind,int ofStatus=0)

{

    if(ind==n)return 0;

    if(dp[ind][ofStatus]!=-1)return dp[ind][ofStatus];

    int r=INT_MIN;

    if(ofStatus!=1)

    {

        r=max(r,a[ind]+res(ind+1,1));

    }

    if(ofStatus!=2)

    {

        r=max(r,b[ind]+res(ind+1,2));

    }

    if(ofStatus!=3)

    {

        r=max(r,c[ind]+res(ind+1,3));

    }

    return dp[ind][ofStatus]=r;

}



int main()

{

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

    cin>>n;

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

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



    cout<<res(0)<<endl;



   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.9630765050988264x):
#include<bits/stdc++.h>

using namespace std;

//#pragma gcc optimize("O3")

//#pragma gcc optimize("unroll-loops")

#ifdef EVAH

    #define _WIN32_WINNT 0x0500

    //#include<windows.h>

#endif

#define openFile(n) ShellExecute(nullptr,"open",n,nullptr,nullptr,SW_SHOWNORMAL);

#define closeP(n) system((string("taskkill /im ")+n+" /f").c_str());

#define closeConsole PostMessage(GetConsoleWindow(), WM_CLOSE, 0, 0);

#define opf "output.txt"

#define ipf "input.txt"

#define mem(n,x) memset(n,x,sizeof(n));

#define sf scanf

#define pf printf

#define ff first

#define ss second

#define pb push_back

#define PII pair<int,int>

#define PLL pair<LL,LL>

#define For(x,n) for(int i=x; i<n; i++)

#define stv v+2*(mid-tl+1)

#define LL long long int

#define N 100001



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



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr); cout.tie(nullptr);

    #ifdef EVAH

        //freopen(ipf,"r",stdin);

        //openFile(ipf);

        //closeConsole;

        //freopen(opf,"w",stdout);

    #endif



    LL n;

    cin >> n;

    vector<LL> one(n),two(n),three(n);

    For(0,n) cin >> one[i] >> two[i] >> three[i];

    dp1[0]=one[0],dp2[0]=two[0],dp3[0]=three[0];

    For(1,n) {

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

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

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

    }

    cout << max(dp1[n-1],max(dp2[n-1],dp3[n-1])) << "\n";



    #ifdef EVAH

        //openFile(opf);

        //Sleep(3000);

        //closeP("notepad.exe");

        //closeConsole;

    #endif

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

    #define ll long long int

    using namespace std;

    int main()

    {

    int a[100000][3]={0};

    int dp[100000][3]={0};

    ll n;

    cin>>n;

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

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

    }



    dp[0][0]=a[0][0];dp[0][1]=a[0][1];dp[0][2]=a[0][2];

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

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

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

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

    }



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

    }


// Faster C++ solution for the same problem (measured speedup: 3.4326756724192746x):
#include <bits/stdc++.h>

#define N 500003

#define MOD 1000000007

using namespace std;





int main()

{

    int n;

    cin>>n;

    int a[n][3];

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

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

            cin>>a[i][j];

    }

    int dp[n][3]={0};

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

        dp[0][i]=a[0][i];

    }

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



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

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

                if(j!=k)

                dp[i][j]=max(a[i][j]+dp[i-1][k],dp[i][j]);

            }

        }

    }

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



}