stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 3.017856166033028x):
#include <bits/stdc++.h>      

using namespace std;

const int N = 1e5 + 100;

#define pb push_back

#define mp make_pair

#define f first

#define s second

#define ull unsigned long long

#define ll long long

#define pii pair <int,int>

#define forn(l,i,n) for(int i = l;i <= n;++i)   

#define FOR(l,i,n) for(int i = l;i < n;++i)

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

#define pll pair <ll,ll>

const ll inf = 1e9 + 7;

const ll ll_inf = 1e17 + 100;   

const double pi = 3.14159265358979323;

inline void fr(){

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

    freopen("gcd.out", "w", stdout);

}

void boost()

{

	ios_base::sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);

}

 

ll binpow (ll a, ll n)

{

    ll res = 1;

    while (n) {

        if (n & 1)

        {

            res *= a;

            res %= inf;

        }

        a *= a;

        n >>= 1;

        a %= inf;

    }

    return res;

}

        

int n;

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

int dp[N][5];

 

main()	

{                                                                        

	cin >> n;

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

	{

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

	}



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

	dp[1][2] = b[1];

	dp[1][3] = c[1];

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

	{

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

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

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

	}

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

}
// 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: 5.054863545826325x):
//by Top1

#include <bits/stdc++.h>

     

#define f first

#define s second

#define int long long

#define ld long double

#define ull unsigned long long

#define pii pair <int, int>

#define pll pair <long, long>

#define mp make_pair

#define pb push_back

#define pf push_front

#define pp pop_back

#define rev reverse

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

#define sz(s) (int)s.size()

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

#define boost ios_base::sync_with_stdio(false); cin.tie(0),cout.tie(0);

#define nxtp next_permutation

#define pvtp prev_permutation

#define KZ return 

#define x1 x21212121

#define y1 y21212121

#define sqr(x) x * x 

 

const int inf = INT_MAX;

const int mod = 1e9 + 7;

const int maxn = 6e5 + 111;

     

using namespace std;



int n, k, a[maxn], b[maxn], c[maxn], dp[5][maxn], q;



void solve(){    

	cin >> n;



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

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



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

	dp[2][1] = b[1];

	dp[3][1] = c[1];



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

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

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

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

	}



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

}



main(){

    boost;

	    

	q = 1;



	while (q--){

		solve();

	}

}
// 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: 2.9824569749212064x):
//by Top1

#include <bits/stdc++.h>

     

#define f first

#define s second

#define int long long

#define ld long double

#define ull unsigned long long

#define pii pair <int, int>

#define pll pair <long, long>

#define mp make_pair

#define pb push_back

#define pf push_front

#define pp pop_back

#define rev reverse

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

#define sz(s) (int)s.size()

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

#define boost ios_base::sync_with_stdio(false); cin.tie(0),cout.tie(0);

#define nxtp next_permutation

#define pvtp prev_permutation

#define KZ return 

#define x1 x21212121

#define y1 y21212121

#define sqr(x) x * x 

 

const int inf = INT_MAX;

const int mod = 1e9 + 7;

const int maxn = 6e5 + 111;

     

using namespace std;



int n, k, a[maxn], b[maxn], c[maxn], dp[5][maxn], q;



void solve(){    

	cin >> n;



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

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



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

	dp[2][1] = b[1];

	dp[3][1] = c[1];



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

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

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

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

	}



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

}



main(){

    boost;

	    

	q = 1;



	while (q--){

		solve();

	}

}
// 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: 5.052296747554921x):
#include <bits/stdc++.h>



#define ll long long

#define f first

#define se second

#define pb push_back

#define mp make_pair

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

#define lb lower_bound

#define ub upper_bound

#define pii pair<int, int>



using namespace std;



const int N = 1e5 + 100, mod = 1e9, inf = 1e9 + 7;

int a[N], b[N], c[N], n, dp[N][4];



ll bin_pow(int x, int y){

    if(y == 0)

        return 1;

    if(y == 1)

        return x;

    if(y % 2 == 0){

        ll z = bin_pow(x, y / 2);

        return (z * z) % mod;

    }

    return (bin_pow(x, y - 1) * x) % mod;

}





int main(){

    boost;

    cin >> n;

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

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

    }

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

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

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

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

    }

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

    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: 2.9809425195183334x):
#include <bits/stdc++.h>



#define ll long long

#define f first

#define se second

#define pb push_back

#define mp make_pair

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

#define lb lower_bound

#define ub upper_bound

#define pii pair<int, int>



using namespace std;



const int N = 1e5 + 100, mod = 1e9, inf = 1e9 + 7;

int a[N], b[N], c[N], n, dp[N][4];



ll bin_pow(int x, int y){

    if(y == 0)

        return 1;

    if(y == 1)

        return x;

    if(y % 2 == 0){

        ll z = bin_pow(x, y / 2);

        return (z * z) % mod;

    }

    return (bin_pow(x, y - 1) * x) % mod;

}





int main(){

    boost;

    cin >> n;

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

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

    }

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

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

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

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

    }

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

    return 0;

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

using namespace std;



const int MaxN = 100005;



int n;

int f[MaxN][3], val[MaxN][3];



void upd(int &a, int b) {

	if (b > a) a = b;

}



int main() {

	scanf("%d", &n);

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

		scanf("%d%d%d", &val[i][0], &val[i][1], &val[i][2]);

	}

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

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

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

				if (j == k) continue;

				upd(f[i][j], f[i - 1][k] + val[i][j]);

			}

		}

	}

	int Ans = 0;

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

			upd(Ans, f[n][j]);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.171067762537888x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

using namespace std;

 

#define mp make_pair

#define pb push_back

#define st first

#define nd second

#define lb lower_bound

#define ub upper_bound

 

typedef long long ll;

typedef long double ld;

typedef pair<int,int> pi;

typedef pair<ll,ll> pll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<pi> vpi;

typedef vector<pll> vpll;

 

const ll oo=1e18;

const int MOD=1000000007;

const int INF=1000000000;

const ld EPS=0.000000001;

 

int dp[100005][3],n,a,b,c;

 

signed main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL);

    cout.tie(NULL);

    cin>>n;

    cin>>a>>b>>c;

    dp[1][0]=a;

    dp[1][1]=b;

    dp[1][2]=c;

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

    }

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.260041677897797x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

using namespace std;

int x[100000][3];

int a[100001][3], mx=0;

int calc(int pos, int u)

{

	if(pos==-1) return 0;

	if(a[pos][u]==-1)

	{

		//if(pos==0) a[pos][u]=mx(x[pos][(u+1)%3],x[pos][(u+2)%3]);

		a[pos][u]=max(calc(pos-1, (u+1)%3)+x[pos][(u+1)%3],calc(pos-1, (u+2)%3)+x[pos][(u+2)%3]);

	}

	return a[pos][u];

}



int main() {

	int n;

	cin>>n;

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

	{

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

		{

			cin>>x[i][u];

		}

	}

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

	{

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

		{

			a[i][u]=-1;

		}

	}

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

	{

		if(mx<calc(n, i) ) mx=calc(n-1, i);

	}

	cout<<mx;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.206438630953253x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

#define LL long long

using namespace std;

const int N =1e6;



LL int dp[N][4], a[N], b[N], c[N];



int main(){

    int n;

    cin >> n;

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

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

    }

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

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

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

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

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

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

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

        //cout << dp[i][0] << " " << dp[i][1] << " " << dp[i][2] << endl;

    }

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

    cout << ans << endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.2159698773201795x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

using namespace std;



int main(){

    int n, i, j, k;

    scanf("%d",&n);

    int a[n+5], b[n+5], c[n+5];

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

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

    }

    int dp[n+5][3];

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

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

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

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

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

    }

    int ans  = 0;

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

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

    cout << ans << "\n";



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.1979148707623475x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

using namespace std;



int main(){



    int n, flag;

    long long ans = 0;



    cin >> n;

    int a[n + 5], b[n + 5], c[n + 5], dp[n + 5][3];

    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];

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

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

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

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

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

    }

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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.208744220453373x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

using namespace std;



typedef long long  int ll;



#define s(t) scanf("%I64d",&t)

#define s2(t,t1) scanf("%I64d%I64d",&t,&t1)





ll n,a[100005][3];

int main()

{

    cin>>n;

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

    {

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

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

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

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



    }

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







}


// Faster C++ solution for the same problem (measured speedup: 5.196729853792935x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

#include <iostream>

#include <string>

#include <vector>

#include <algorithm>

#include <sstream>

#include <queue>

#include <deque>

#include <bitset>

#include <iterator>

#include <list>

#include <stack>

#include <map>

#include <set>

#include <functional>

#include <numeric>

#include <utility>



#include <cmath>

#include <iomanip>



using namespace std;



int main()

{

  ios::sync_with_stdio(false);

  cin.tie(0);

  int n;

  cin >> n;

  int dp[n][3]; // dp[i][j] represents the maximum happiness at the end of day i+1 if Taro did activity number j that day.

  int cur;

  cin >> cur;

  dp[0][0] = cur;

  cin >> cur;

  dp[0][1] = cur;

  cin >> cur;

  dp[0][2] = cur;

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

    cin >> cur;

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

    cin >> cur;

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

    cin >> cur;

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

  }

  cout << *max_element(dp[n-1], dp[n-1]+3) << "\n";

  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.26633937790592x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

using namespace std;

#define ff first

#define ss second

#define ll long long int

#define pii pair<int,int>

#define pll pair<ll,ll>

#define mem(x) memset(x, 0, sizeof x)

#define mem1(x) memset(x, -1, sizeof x)

#define MX 100005

ll arr[MX][3],arr2[MX],arr3[MX], dp[MX][3],n;

//ll rec(ll i, ll score)

//{

//    if(i==n-1) return score;

//    else if(i>=n) return 1e9;

//    else if(dp[i]<score && dp[i]!=-1) return dp[i];

//    else if(!i || dp[i]==-1 || score<dp[i])

//    {

//        ll ret1=1e9, ret2=1e9;

//        if(i+1<=n-1) ret1= rec(i+1, score+llabs(arr[i+1]-arr[i]));

//        if(i+2<=n-1) ret2= rec(i+2, score+llabs(arr[i+2]-arr[i]));

//        return dp[i]=min(ret1,ret2);

//    }

//}

int main()

{

    ll i,j,k,l, m;

    scanf("%lld",&n);

    for(i=0; i<n; i++) scanf("%lld%lld%lld",&arr[i][0],&arr[i][1],&arr[i][2]);

//    for(i=1; i<n; i++) dp[i]=1e9;

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

    {

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

        {

            ll mx=0;

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

                if(k!=j) mx=max(arr[i-1][k],mx);

            arr[i][j]+=mx;

        }

    }

    printf("%lld\n",max(arr[n-1][0], max(arr[n-1][1],arr[n-1][2])));

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.130695417210999x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

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

using namespace std;

int n;

int a[110000][4];

int m[110000][4];

int dp(int numTaken, int prevTaken){

	if(numTaken == n+1){

		return 0;

	}

	

	if(m[numTaken][prevTaken]!=-1) return m[numTaken][prevTaken];

	

	int ans = 0;

	// 3 pilihan

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

		if(i != prevTaken){

			ans = max(ans, dp(numTaken+1, i) + a[numTaken][i-1]);

		}

	}

	return m[numTaken][prevTaken] = ans;

}

int main(){

	cin>>n;

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

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

			cin>>a[i][j];

		}

	}

	

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

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

			m[i][j] = -1;

		}

	}

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

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

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

//			cout<<m[i][j]<<" ";

//		}cout<<"\n";

//	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.21010836138604x):
#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int n,a[3][100005],f[3][100005];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[0][i],&a[1][i],&a[2][i]);

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

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

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

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

	}

	printf("%d\n",max(f[0][n],max(f[1][n],f[2][n])));

	return 0;

}