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

using namespace std;



int m;

const int mx = 1e5+5;

int tasks[mx][3];

int diario[3][mx];

int cosa(int a, int i){

	if(i == m){

     return 0;

     }



	if(diario[a][i] != -1) {

        return diario[a][i];

    }

	int& ans = diario[a][i];



        if(a==0){

		ans=max(ans, cosa(1, i+1)+tasks[i][0]);

		ans=max(ans, cosa(2, i+1)+tasks[i][0]);

	}

        if(a==1){

		ans=max(ans, cosa(0, i+1)+tasks[i][1]);

		ans=max(ans, cosa(2, i+1)+tasks[i][1]);

        }

	if(a==2){

		ans=max(ans, cosa(0, i+1)+tasks[i][2]);

		ans=max(ans, cosa(1, i+1)+tasks[i][2]);

	}



	return ans;

}

int main(){

	cin>>m;

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

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

	}

	memset(diario, -1, sizeof diario);

	cout<<max(cosa(0,0),max(cosa(1,0),cosa(2,0)))<<"\n";

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.208573480884606x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

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

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

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

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

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

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

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

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

    }

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

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	return 0;

}
// 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;



int nn;

const int nxx = 1e5+5;

int acc[nxx][3];

int dpp[3][nxx];

int go(int aa, int ii)

{

	if(ii==nn) return 0;

	if(dpp[aa][ii] != -1) return dpp[aa][ii];

	int& ans = dpp[aa][ii];

	if(aa==0)

	{

		ans=max(ans, go(1, ii+1)+acc[ii][0]);

		ans=max(ans, go(2, ii+1)+acc[ii][0]);

	}

	if(aa==1)

	{

		ans=max(ans, go(0, ii+1)+acc[ii][1]);

		ans=max(ans, go(2, ii+1)+acc[ii][1]);

	}

	if(aa==2)

	{

		ans=max(ans, go(0, ii+1)+acc[ii][2]);

		ans=max(ans, go(1, ii+1)+acc[ii][2]);

	}



	return ans;



}

int main()

{

    cin>>nn;

    for(int ii=0; ii<nn; ii++) cin>>acc[ii][0]>>acc[ii][1]>>acc[ii][2];

    memset(dpp, -1, sizeof dpp);

    cout<<max(go(0,0), max(go(1,0),go(2,0)))<<"\n";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2085712629077756x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

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

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

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

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

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

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

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

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

    }

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

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	return 0;

}
// 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;



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: 2.208547160708751x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

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

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

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

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

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

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

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

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

    }

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

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	return 0;

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

using namespace std;

int v[100009][3];

int dp[100009][3];

int n;

int vacation(int i, int j)

{

    if(i==n-1)

        return dp[i][j]=v[i][j];

    if(i>=n || j>=3)

        return 0;

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

        return dp[i][j];

    int temp = INT_MIN;

        if(j==0)

        {

             temp = max(vacation(i+1,j+1)+v[i][j],temp);

        temp = max(vacation(i+1,j+2)+v[i][j],temp);

 return dp[i][j]= temp;

        }

      else  if(j==1)

        {

             temp = max(vacation(i+1,0)+v[i][j],temp);

        temp = max(vacation(i+1,2)+v[i][j],temp);

 return dp[i][j]= temp;

        }

        else

        {

             temp = max(vacation(i+1,1)+v[i][j],temp);

        temp = max(vacation(i+1,0)+v[i][j],temp);

         return dp[i][j]= temp;

        }

}



int main()

{

    cin>>n;

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

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

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

    cout<<max(vacation(0,0),max(vacation(0,1),vacation(0,2)))<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 2.2086691687691826x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

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

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

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

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

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

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

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

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

    }

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

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	return 0;

}
// 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: 2.2189849292866173x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

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

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

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

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

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

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

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

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

    }

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

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	return 0;

}
// 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: 5.104203716846349x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

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

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

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

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

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

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

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

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

    }

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

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	return 0;

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

#include <algorithm>

#include <cstring>

const int maxn = 3e5 + 5;

using namespace std;



int N;

int a[maxn][4];

int dp[maxn][4];

int main()

{

	scanf("%d", &N);

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

	{

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

		{

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

		}

	}

	memset(dp, 0, sizeof dp);

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

		dp[1][i] = a[1][i];

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

	{

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

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

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

	}

	int ans = 0;

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

		ans = max(dp[N][i], ans);

	printf("%d\n", ans);

	return 0;	

}

// Faster C++ solution for the same problem (measured speedup: 1.7103694262178617x):
#include <iostream>

#include <queue>

#include <string>

#include <map>

#include <stdio.h>

#include <iomanip>

#include <math.h>

#include <vector>

#include <stack>

#include <algorithm>

#include <string.h>

#include <set>

#include <cstdlib>

#include <iostream>

#include <bits/stdc++.h>

#define pii pair<int,int> 

#define ll long long int 

#define sl(n) scanf("%lld",&n)

#define s(n) scanf("%d",&n)

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

#define fo(i,n) for(int i=1;i<=n;++i)

#define fnm(n,m) for(ll i=n;i<m+1;++i)

#define pb push_back

#define pf printf

#define F first

#define S second

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

#define srtc(v,cmp) sort(v.begin(),v.end(),cmp);

#define srta(v,n) sort(v,v+n);

const ll LL_INF=0x3f3f3f3f3f3f;

const ll MOD=1073741824;





using namespace std;





int T;

int d,arr[100005][5];

int dp[100005][5];



int gen(int i,int prev){ // swim:0, bugs:1, homework:2

    if(i == d)return 0;

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

    if(~ret)return ret;

    ret = gen(i+1,(prev+1)%3)+arr[i][(prev+1)%3];

    ret = max( ret , gen(i+1,(prev+2)%3)+arr[i][(prev+2)%3]);

    if(!i)ret = max( ret , gen(i+1,(prev)%3)+arr[i][(prev)%3]);

    return ret;

}



int main(){

    // pf("\t\t\t***************************************************************************\n\n");

    // pf("\t\t\t***************************************************************************\n\n");

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

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

    memset(dp,-1,sizeof dp);

    s(d);

    f(i,d){

        s(arr[i][0]);

        s(arr[i][1]);

        s(arr[i][2]);

    }

    pf("%d\n",gen(0,0));



    return 0;

}
// 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: 1.7071901151200943x):
#include <iostream>

#include <queue>

#include <string>

#include <map>

#include <stdio.h>

#include <iomanip>

#include <math.h>

#include <vector>

#include <stack>

#include <algorithm>

#include <string.h>

#include <set>

#include <cstdlib>

#include <iostream>

#include <bits/stdc++.h>

#define pii pair<int,int> 

#define ll long long int 

#define sl(n) scanf("%lld",&n)

#define s(n) scanf("%d",&n)

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

#define fo(i,n) for(int i=1;i<=n;++i)

#define fnm(n,m) for(ll i=n;i<m+1;++i)

#define pb push_back

#define pf printf

#define F first

#define S second

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

#define srtc(v,cmp) sort(v.begin(),v.end(),cmp);

#define srta(v,n) sort(v,v+n);

const ll LL_INF=0x3f3f3f3f3f3f;

const ll MOD=1073741824;





using namespace std;





int T;

int d,arr[100005][5];

int dp[100005][5];



int gen(int i,int prev){ // swim:0, bugs:1, homework:2

    if(i == d)return 0;

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

    if(~ret)return ret;

    ret = gen(i+1,(prev+1)%3)+arr[i][(prev+1)%3];

    ret = max( ret , gen(i+1,(prev+2)%3)+arr[i][(prev+2)%3]);

    if(!i)ret = max( ret , gen(i+1,(prev)%3)+arr[i][(prev)%3]);

    return ret;

}



int main(){

    // pf("\t\t\t***************************************************************************\n\n");

    // pf("\t\t\t***************************************************************************\n\n");

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

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

    memset(dp,-1,sizeof dp);

    s(d);

    f(i,d){

        s(arr[i][0]);

        s(arr[i][1]);

        s(arr[i][2]);

    }

    pf("%d\n",gen(0,0));



    return 0;

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

#include <algorithm>

#include <cstring>

const int maxn = 3e5 + 5;

using namespace std;



int N;

int a[maxn][4];

int dp[maxn][4];

int main()

{

	scanf("%d", &N);

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

	{

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

		{

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

		}

	}

	memset(dp, 0, sizeof dp);

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

		dp[1][i] = a[1][i];

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

	{

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

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

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

	}

	int ans = 0;

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

		ans = max(dp[N][i], ans);

	printf("%d\n", ans);

	return 0;	

}

// Faster C++ solution for the same problem (measured speedup: 5.11105206450666x):
#include <iostream>

#include <algorithm>



int arr[100005][3];

int val[100005][3];



int main()

{

	int n;

	std::cin >> n;

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

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

	val[0][0] = arr[0][0];

	val[0][1] = arr[0][1];

	val[0][2] = arr[0][2];

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

	{

		int* curr = val[i];

		int* temp = val[i - 1];

		curr[0] = std::max(temp[1], temp[2]) + arr[i][0];

		curr[1] = std::max(temp[0], temp[2]) + arr[i][1];

		curr[2] = std::max(temp[0], temp[1]) + arr[i][2];

	}

	std::cout << std::max({ val[n - 1][0], val[n - 1][1], val[n - 1][2] }) << '\n';

	return 0;

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

using namespace std;

#define db(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<'\n'

const int N = 1e5;

int A[N+7][5], mem[N+7][5];



int dp(int i, int j) {

	if(i==0) return 0;

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

	if(j==1) return mem[i][j] = A[i][j] + max(dp(i-1, 2), dp(i-1, 3));

	if(j==2) return mem[i][j] = A[i][j] + max(dp(i-1, 3), dp(i-1, 1));

	else return mem[i][j] = A[i][j] + max(dp(i-1, 1), dp(i-1, 2));

}

int main(){

	ios_base::sync_with_stdio(0);

	int a,b,n,T,cas=0;

	cin>>n;

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

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

	}

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

	int res = max(dp(n,1), max(dp(n,2), dp(n,3)));

	cout<<res<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.0156091733790706x):
#include <iostream>

#include <algorithm>



int arr[100005][3];

int val[100005][3];



int main()

{

	int n;

	std::cin >> n;

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

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

	val[0][0] = arr[0][0];

	val[0][1] = arr[0][1];

	val[0][2] = arr[0][2];

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

	{

		int* curr = val[i];

		int* temp = val[i - 1];

		curr[0] = std::max(temp[1], temp[2]) + arr[i][0];

		curr[1] = std::max(temp[0], temp[2]) + arr[i][1];

		curr[2] = std::max(temp[0], temp[1]) + arr[i][2];

	}

	std::cout << std::max({ val[n - 1][0], val[n - 1][1], val[n - 1][2] }) << '\n';

	return 0;

}
// Slower C++ solution:
/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



int n;

int a[N][3];

int mem[N][3];



int go(int idx , int lst){

    if(idx == n + 1)

        return 0;

    if(~mem[idx][lst])

        return mem[idx][lst];

    int best = 0;

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

        if(idx == 1 || lst != i){

            best = max(best , go(idx + 1 , i) + a[idx][i]);

        }

    }

    return mem[idx][lst] = best;

}



int main(){

    scanf("%d" , &n);

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

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

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

        }

    }

    memset(mem , -1 , sizeof mem);

    printf("%d\n" , go(1 , 0));

}


// Faster C++ solution for the same problem (measured speedup: 2.188785519295095x):
#include <iostream>

#include <algorithm>



int arr[100005][3];

int val[100005][3];



int main()

{

	int n;

	std::cin >> n;

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

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

	val[0][0] = arr[0][0];

	val[0][1] = arr[0][1];

	val[0][2] = arr[0][2];

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

	{

		int* curr = val[i];

		int* temp = val[i - 1];

		curr[0] = std::max(temp[1], temp[2]) + arr[i][0];

		curr[1] = std::max(temp[0], temp[2]) + arr[i][1];

		curr[2] = std::max(temp[0], temp[1]) + arr[i][2];

	}

	std::cout << std::max({ val[n - 1][0], val[n - 1][1], val[n - 1][2] }) << '\n';

	return 0;

}
// 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: 2.2073995303755725x):
#include <iostream>

#include <algorithm>



int arr[100005][3];

int val[100005][3];



int main()

{

	int n;

	std::cin >> n;

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

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

	val[0][0] = arr[0][0];

	val[0][1] = arr[0][1];

	val[0][2] = arr[0][2];

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

	{

		int* curr = val[i];

		int* temp = val[i - 1];

		curr[0] = std::max(temp[1], temp[2]) + arr[i][0];

		curr[1] = std::max(temp[0], temp[2]) + arr[i][1];

		curr[2] = std::max(temp[0], temp[1]) + arr[i][2];

	}

	std::cout << std::max({ val[n - 1][0], val[n - 1][1], val[n - 1][2] }) << '\n';

	return 0;

}