stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#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));





}

// Faster C++ solution for the same problem (measured speedup: 2.559434862942615x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

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

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int N = 1e5+10 , OO = 0x3f3f3f3f ;

//const int mod = 1000000007;

 

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

double startTime;

double getCurrentTime() {

	return ((double)clock() - startTime) / CLOCKS_PER_SEC;

}

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

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

#define rloop(i,n) for(int i = int(n)-1;i >= 0;i--)

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

#define range(vec) for(auto &x : vec) cin >> x;

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

#define SZ(v)  ((int)((v).size()))

int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction

//int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 direction

//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction

//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction

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

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



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

int dp[N][4];



int solve(int idx , int lst){

	

	if(idx == n) return 0;

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

	if(~ret) return ret;

	ret = -OO;

	if(lst != 0) // swim

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

	if(lst != 1)

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

	if(lst != 2)

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

		

	return ret;

}





int main()

{

    startTime = (double)clock();

    #ifndef ONLINE_JUDGE

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

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

    #endif

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

	

	cin >> n;

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

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



	clr(dp , -1);

	cout << solve(0 , 3) << "\n";





    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.5993754024847715x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

}
// Slower C++ solution:
#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);

}


// Faster C++ solution for the same problem (measured speedup: 2.9828872094488865x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 2.960951533907414x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

}
// Slower C++ solution:
#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();

}


// Faster C++ solution for the same problem (measured speedup: 2.5876780260556775x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

}
// 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 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;

}

// Faster C++ solution for the same problem (measured speedup: 2.5885161460815778x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

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





using namespace std;



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

int n;



long long meme[100007][5];





long long solve(int index, int flag){

    if(index==n){

        return 0;

    }

    

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

    

    if(res != -1){

        return res;

    }

    res = 0;

    

    if(flag!=0){

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

    }

    if(flag!=1){

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

    }

    if(flag!=2){

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

    }

    return res;

}



int main(){

    cin>>n;

    int num1,num2,num3;

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

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

        cin>>num1>>num2>>num3;

        vec[i].push_back(num1);

        vec[i].push_back(num2);

        vec[i].push_back(num3);

    }

    long long sol = solve(0,3);

    cout<<sol<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 7.968955524445398x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

}
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 2.2093150256898664x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

}
// Slower C++ solution:
/**

 * @author      : Luis Miguel Baez <es.luismiguelbaez@gmail.com>

 * @created     : Thursday August 13, 2020

 */



#include <bits/stdc++.h>



using namespace std;



#define endl '\n'

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

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

#define ms(arr, value) memset(arr, value, sizeof arr)

#define ll    int64_t



#define pii   pair<int, int>

#define piii  tuple<int, int, int>

#define pll   pair<ll, ll>

#define plll  tuple<ll, ll, ll>



#define F  first

#define S  second



const int oo  = (int)(1e9);

const ll lloo  = (ll)(1e18);



const int N = int(1e5) + 10;



class Task {

public:



    int n;

    vector<piii> A;



    ll dp[N][4];





    ll go(int index, int last) {

        if(index > n) return 0LL;

        ll &ans = dp[index][last];

        if(ans != -1) return ans;

        ll x=0, y=0;

        if(last == 1) {

            x = get<1>(A[index]) + go(index+1, 2);

            y = get<2>(A[index]) + go(index+1, 3);

        } else if(last == 2) {

            x = get<0>(A[index]) + go(index+1, 1);

            y = get<2>(A[index]) + go(index+1, 3);

        } else if(last == 3) {

            x = get<0>(A[index]) + go(index+1, 1);

            y = get<1>(A[index]) + go(index+1, 2);

        } else {

            ll z = 0;

            x = get<0>(A[index]) + go(index+1, 1);

            y = get<1>(A[index]) + go(index+1, 2);

            z = get<2>(A[index]) + go(index+1, 3);

            x = max(x, z);

        }

        ans = max(x, y);

        return ans;

    }



    ll solveOne(istream &in, ostream &out) {

        int a, b, c;

        in >> n;

        A.resize(n+1);

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

            in >> a >> b >> c;

            A[i] = {a, b, c};

        }

        memset(dp, -1, sizeof dp);

        ll ans = go(1, -1);

        return ans;  

    }

    

    void solve(istream &in, ostream &out) {

        int t = 1;

        while(t--) {

            out << solveOne(in, out) << endl;

        }

    }

};



auto main() -> int {

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    

    Task solver;

    istream &in(cin);

    ostream &out(cout);

    solver.solve(in, out);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.235973839157217x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

}
// Slower C++ solution:
#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';

}

// Faster C++ solution for the same problem (measured speedup: 2.1814815503090417x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int main(){

    ll n;

    cin >> n;

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

    ll i,j;

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

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

    }

    vector<vector<ll>> dp(n,vector<ll>(3,0));

    

    // base case;

    // in first day he can do anything his wish;

    

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    

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

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

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

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

    }

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

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

using namespace std;

#define print printf("==================\n")

#define ll long long

#define pi acos(-1.0)

#define eps 1e-16

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

const ll INF = 1 << 30;

typedef pair<ll, ll  > payar;

typedef struct

{

    ll  x, y;

} point;

priority_queue<payar, vector<payar>,  greater<payar> > pq; ///accending

vector< pair<ll,payar  > > vpp;

vector < payar >  vp;



ll n ;

int ara[100005][4];

ll dp[100005][4];

ll cal(int day, int itm)

{

    if(day>n)

    {

        return 0;

    }



    if(dp[day][itm]!=-1)

        return dp[day][itm];

    ll maxi = -1;

    if(day==1)

    {

   return   maxi = dp[day][itm]=max3( cal(day+1,1)+ara[day][1], cal(day+1,2)+ara[day][2], cal(day+1,3)+ara[day][3] );



    }

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

    {

            if(i==itm)

                continue;

            maxi=max( cal(day+1,i)+ara[day][i],maxi);





    }

    return dp[day][itm]=maxi;

}



int main()

{



    cin>>n;

    memset(dp,-1,sizeof dp);

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

    {

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

        {

            cin>>ara[i][j];

        }

    }

    cout<<cal(1,1);







    return  0 ;

}


// Faster C++ solution for the same problem (measured speedup: 1.9092586160892964x):
#include<bits/stdc++.h>

using namespace std;

int ara[100005][3];

int dp[100005][3];



int func(int pos,int des,int idx){

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

    int res;

    if(pos>des) res=0;

    else{

        int t1,t2;

        if(idx==0){

            t1=ara[pos][0]+func(pos+1,des,1);

            t2=ara[pos][0]+func(pos+1,des,2);

        }

        else if(idx==1){

             t1=ara[pos][1]+func(pos+1,des,0);

            t2=ara[pos][1]+func(pos+1,des,2);

        }

        else{

             t1=ara[pos][2]+func(pos+1,des,1);

            t2=ara[pos][2]+func(pos+1,des,0);

        }

        res=max(t1,t2);

    }

    dp[pos][idx]=res;

    return res;

}

int main(){

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

    int n,x,y,z;

    cin >> n;

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

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

    }

    x=func(1,n,0);

    y=func(1,n,1);

    z=func(1,n,2);

    int ans;

    

    ans=max(x,y);

    ans=max(ans,z);

    

    cout << ans << endl;

    

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

using namespace std;

#define N (int)2e5

int n,a[N],b[N],c[N],memo[N][10];



// a = 0, b = 1, c = 2

int dp(int curr , int last) {

    if (curr > n) return 0;

    int &sol = memo[curr][last];

    if (sol != -1) return sol;

    if (last == 0) {

        sol = max(

            dp(curr + 1 , 1) + b[curr] ,

            dp(curr + 1 , 2) + c[curr]

        );

    } else if (last == 1) {

        sol = max(

            dp(curr + 1 , 0) + a[curr] ,

            dp(curr + 1 , 2) + c[curr]

        );

    } else {

        sol = max(

            dp(curr + 1 , 0) + a[curr] ,

            dp(curr + 1 , 1) + b[curr]

        );

    }

    return sol;

}



int main() {

    scanf("%d" , &n);

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

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

    memset (memo , -1 , sizeof memo);

    printf("%d\n" , max(dp(2 , 0) + a[1] , max(dp(2 , 1) + b[1] , dp(2 , 2) + c[1])));

}


// Faster C++ solution for the same problem (measured speedup: 4.235652719014905x):
#include<bits/stdc++.h>

using namespace std;

int ara[100005][3];

int dp[100005][3];



int func(int pos,int des,int idx){

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

    int res;

    if(pos>des) res=0;

    else{

        int t1,t2;

        if(idx==0){

            t1=ara[pos][0]+func(pos+1,des,1);

            t2=ara[pos][0]+func(pos+1,des,2);

        }

        else if(idx==1){

             t1=ara[pos][1]+func(pos+1,des,0);

            t2=ara[pos][1]+func(pos+1,des,2);

        }

        else{

             t1=ara[pos][2]+func(pos+1,des,1);

            t2=ara[pos][2]+func(pos+1,des,0);

        }

        res=max(t1,t2);

    }

    dp[pos][idx]=res;

    return res;

}

int main(){

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

    int n,x,y,z;

    cin >> n;

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

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

    }

    x=func(1,n,0);

    y=func(1,n,1);

    z=func(1,n,2);

    int ans;

    

    ans=max(x,y);

    ans=max(ans,z);

    

    cout << ans << endl;

    

}