stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include<iostream>

#include<cstring>

using namespace std;



#define Max 100001

int dp[Max][3];



int main(){

	int n,a,b,c;

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

	cin>>n;

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

		cin>>a>>b>>c;

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

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

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

	}

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

	cout<<ans<<endl;

	return 0;

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std ;

#define ll long long

inline int D(){

    int t ;

    scanf( "%d" , &t ) ;

    return t ;

}

inline ll LLD(){

    ll t ;

    scanf( "%lld" , &t ) ;

    return t ;

}

//################################################################

const int N=100009;

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

int mem[N][4];

int solve(int ind ,int lastInd){

    if(ind==n)return 0;

    int  ret=INT_MIN;

    if(mem[ind][lastInd]!=-1)return mem[ind][lastInd];

    if(lastInd!=0){



        ret=max(ret,a[ind]+solve(ind+1,0));

    }

    if(lastInd!=1){



        ret=max(ret,b[ind]+solve(ind+1,1));

    }

    if(lastInd!=2){



        ret=max(ret,c[ind]+solve(ind+1,2));

    }

    return mem[ind][lastInd]=ret;



}

int main() {

    n=D();

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

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

    }

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

    printf("%d",solve(0,3));





}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define F first

#define S second

#define T second.first

#define C second.second

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

using namespace std;

typedef long long ll;

typedef pair<int, int> pii;

typedef pair<int,pair<int,int>> piii;



const int N= 1e5+5, L = 14, OO = 0x3f3f3f3f;

const ll M=10 ;

const double ESP = 1e-8;



int n, k;

piii A[N];

int dp[N][5];

int solve(int idx, int prev){

    if(idx == n)

        return 0;



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

    if(~ret)

        return ret;



    ret = max(ret, (prev != 1)? solve(idx+1, 1)+A[idx].F : -1);

    ret = max(ret, (prev != 2)? solve(idx+1, 2)+A[idx].T : -1);

    return ret = max(ret, (prev != 3)? solve(idx+1, 3)+A[idx].C : -1);



}



int main(){

    FIO

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

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

    cin>>n;

    for(int i=0; i<n; i++)cin>>A[i].F>>A[i].T>>A[i].C;

    memset(dp, -1, sizeof dp);

    cout<<solve(0,4);

}


// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>



using namespace std;



const int N=1e5;

int n;

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

int mem[N][5];



int solve(int i,int lst){

    if(i==n)

        return 0;

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

    int op1=lst!=0?solve(i+1,0)+a[i]:-1000000000;

    int op2=lst!=1?solve(i+1,1)+b[i]:-1000000000;

    int op3=lst!=2?solve(i+1,2)+c[i]:-1000000000;

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

}



int main() {

    scanf("%d",&n);

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

        scanf("%d%d%d",a+i,b+i,c+i);

    memset(mem,-1,sizeof mem);

    printf("%d",max(solve(0,0),max(solve(0,1),solve(0,2))));

    return 0;

}


// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define ll long long

#define ull unsigned long long

#define PII pair<int,int>

#define PSI pair<string,int>

#define PIII pair< int, pair<int,int> >

const int MXN = 1e5+5;

int a[MXN][4], dp[MXN][4];

int N;



int rec(int pos, int type){



    if(pos>N) return 0;



    if(dp[pos][type]!=-1) return dp[pos][type];



    int ret = -1;

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

        if(i!=type)

            ret = max (ret, rec(pos+1,i) + a[pos+1][i]);



    return dp[pos][type] = ret;

}



void read(){



    int i,j;

    cin>>N;

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

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

            cin>>a[i][j];

}



void solve(){



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

    read();

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

}



int main(){



    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);

    solve();

}


// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

int m[MXN][3];

int dp[MXN][4];



int x[3] = {0, 1, 2};



int Dp(int d, int lst) {

    if(d == n) return 0;

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

    int ans = 0;

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

        if(lst != x[i]) {

            ans = max(ans, Dp(d + 1, i) + m[d][i]);

        }

    }

    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 >> m[i][0] >> m[i][1] >> m[i][2];

    }

    memset(dp, -1, sizeof dp);

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

    return 0;

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

//#define int long long

#define fast ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)

using namespace std;

#define endl '\n'

#define fori(s,n) for(int i=s;i<n;i++) 

#define forj(s,n) for(int j=s;j<n;j++) 

#define vii vector<int>

#define pii pair<int,int>

int n;

vector<vii> v;

int maxi = 0;

int dp[100005][3];

int solve(int x, int y) {//{0,0} intially

	

	if (x == n) {

		return 0;

	}

	int& ret = dp[x][y];

	if (ret != -1) return ret;

	if (y != 0 || x == 0)

		ret = max(ret, v[x][0] + solve(x + 1, 0));

	if (y != 1 || x == 0)

		ret = max(ret, v[x][1] + solve(x + 1, 1));

	if (y != 2 || x == 0)

		ret = max(ret, v[x][2] + solve(x + 1, 2));

	return ret;

}

signed main()

{

	fast;

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

	cin >> n;

	v = vector<vii>(n, vii(3));

	fori(0, n)

		forj(0, 3)

		cin >> v[i][j];

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

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

int k;

const int N=1e5+10;

int f[N][3];

int memo[N][3];

 

int dp(int pos, int now){

    if(pos==k) return 0;

    if(memo[pos][now]!=-1) return memo[pos][now];

    else{

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

            if(now==i) continue;

            else{

            memo[pos][now]=max(memo[pos][now],dp(pos+1,i)+f[pos][i]);

            }

        }

    }

    return memo[pos][now];

    

}

 

int main(){

    cin>>k;

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

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

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

            cin>>f[i][j];

        }

    }

    int ans=0;

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

        ans=max(ans,dp(0,i));

    }

    cout<<ans<<'\n';

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



#define FASTREAD      ios_base::sync_with_stdio(0);cin.tie(nullptr);

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

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

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

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

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

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

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

    ptl(ans);

    return 0;

}


// Fastest known solution for this problem: