stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

#define format(n) fixed<<setprecision(n)

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

#define clr(dp,n) memset(dp,n,sizeof dp)

#define ex(n) return cout<<n<<finl,0;

int dx[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

int dy[] = { 1, 0, -1, 0, 1, -1, 1, -1 };

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

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





}

int n;

vector<vector<ll>>v;

ll mem[(int)1e5 + 10][5];

ll solve(int i, int c)

{

	if (i == n)return 0;

	ll& ret = mem[i][c];

	if (~ret)return ret;

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

	{

		if (k == c)continue;

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

	}

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		cin >> n;

	v = vector<vector<ll>>(n + 1, vector<ll>(3));

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

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

	clr(mem, -1);

	ex(solve(0, 4));

}

// Faster C++ solution for the same problem (measured speedup: 4.86411690651849x):
#include<bits/stdc++.h>



using namespace std;

const int N=150000;

int main(){

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

	int i,n,A,B,C,a[N],b[N],c[N];

	cin>>n;

	cin>>a[0]>>b[0]>>c[0];

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

		cin>>A>>B>>C;

		a[i]=max(b[i-1],c[i-1])+A;

		b[i]=max(a[i-1],c[i-1])+B;

		c[i]=max(b[i-1],a[i-1])+C;

		

	}

	cout<<max(a[n-1],max(b[n-1],c[n-1]));

}
// 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: 1.953147478527672x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

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

int dp[MXN][4];



int Dp(int d, int lst) {

    if(d == n) return 0;

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

    int ans = 0;

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

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

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

    return dp[d][lst] = ans;

}



int main() {

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

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

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

    return 0;

}
// 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: 1.152390814821169x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

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

int dp[MXN][4];



int Dp(int d, int lst) {

    if(d == n) return 0;

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

    int ans = 0;

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

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

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

    return dp[d][lst] = ans;

}



int main() {

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

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

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

    return 0;

}
// 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: 1.9495168807405079x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

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

int dp[MXN][4];



int Dp(int d, int lst) {

    if(d == n) return 0;

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

    int ans = 0;

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

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

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

    return dp[d][lst] = ans;

}



int main() {

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

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

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

    return 0;

}
// Slower C++ solution:
#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;

} 

// Faster C++ solution for the same problem (measured speedup: 1.1419448036127873x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

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

int dp[MXN][4];



int Dp(int d, int lst) {

    if(d == n) return 0;

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

    int ans = 0;

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

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

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

    return dp[d][lst] = ans;

}



int main() {

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

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

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

    return 0;

}
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

#define format(n) fixed<<setprecision(n)

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

#define clr(dp,n) memset(dp,n,sizeof dp)

#define ex(n) return cout<<n<<finl,0;

int dx[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

int dy[] = { 1, 0, -1, 0, 1, -1, 1, -1 };

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

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





}

int n;

vector<vector<ll>>v;

ll mem[(int)1e5 + 10][5];

ll solve(int i, int c)

{

	if (i == n)return 0;

	ll& ret = mem[i][c];

	if (~ret)return ret;

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

	{

		if (k == c)continue;

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

	}

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		cin >> n;

	v = vector<vector<ll>>(n + 1, vector<ll>(3));

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

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

	clr(mem, -1);

	ex(solve(0, 4));

}

// Faster C++ solution for the same problem (measured speedup: 1.9507081411661729x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

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

int dp[MXN][4];



int Dp(int d, int lst) {

    if(d == n) return 0;

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

    int ans = 0;

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

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

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

    return dp[d][lst] = ans;

}



int main() {

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

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

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

    return 0;

}
// 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: 2.315922229698524x):
#include <bits/stdc++.h>



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

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

        return dp[y][x];

    if(y == 0 )

    {

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

        return A[y][x];

    }

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

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

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

}
// 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: 1.3664341964369122x):
#include <bits/stdc++.h>



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

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

        return dp[y][x];

    if(y == 0 )

    {

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

        return A[y][x];

    }

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

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

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

}
// 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: 2.311617289997439x):
#include <bits/stdc++.h>



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

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

        return dp[y][x];

    if(y == 0 )

    {

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

        return A[y][x];

    }

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

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

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

}
// Slower C++ solution:
#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;

} 

// Faster C++ solution for the same problem (measured speedup: 1.3540479584107865x):
#include <bits/stdc++.h>



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

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

        return dp[y][x];

    if(y == 0 )

    {

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

        return A[y][x];

    }

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

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

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

}
// Slower C++ solution:


#include <bits/stdc++.h>

using namespace  std;



int N;

const int SIZE = 100005;

int mem [SIZE][4];

int arr [SIZE][3];





int dp (int i, int activ) {

  

  if (i == N) return 0;

  



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

  

  int ret = -1; 



  if (activ != 0) {

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

  }



  if (activ != 1) {

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

    

  }



  if (activ != 2) {

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

    

  }



  return  mem[i][activ] = ret;

}



int main (){ 



  cin  >> N ;

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

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

  }



  memset(mem, -1, sizeof mem);

  





  cout << dp(0,3);



}

// Faster C++ solution for the same problem (measured speedup: 1.1791585715385915x):
#include <bits/stdc++.h>



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

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

        return dp[y][x];

    if(y == 0 )

    {

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

        return A[y][x];

    }

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

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

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

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



using namespace std;



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

int memo[100007][4];

int n; 



int solve(int p, int last) {

    if(p == n) return 0;

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

    if(st == -1) {

        st = max({

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

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

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

        });

    }

    return st;

}

int main() {

    memset(memo, -1, sizeof memo);

    cin >> n;

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

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

    }



    cout << solve(0, 0);

}

// Faster C++ solution for the same problem (measured speedup: 1.1820515520125598x):
#include <bits/stdc++.h>



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

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

        return dp[y][x];

    if(y == 0 )

    {

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

        return A[y][x];

    }

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

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

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

}
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

#define format(n) fixed<<setprecision(n)

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

#define clr(dp,n) memset(dp,n,sizeof dp)

#define ex(n) return cout<<n<<finl,0;

int dx[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

int dy[] = { 1, 0, -1, 0, 1, -1, 1, -1 };

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

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





}

int n;

vector<vector<ll>>v;

ll mem[(int)1e5 + 10][5];

ll solve(int i, int c)

{

	if (i == n)return 0;

	ll& ret = mem[i][c];

	if (~ret)return ret;

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

	{

		if (k == c)continue;

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

	}

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		cin >> n;

	v = vector<vector<ll>>(n + 1, vector<ll>(3));

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

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

	clr(mem, -1);

	ex(solve(0, 4));

}

// Faster C++ solution for the same problem (measured speedup: 2.3130298133892913x):
#include <bits/stdc++.h>



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

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

        return dp[y][x];

    if(y == 0 )

    {

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

        return A[y][x];

    }

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

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

}

int main()

{

    int bas;

    cin>>n;

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

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

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

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

    	bas = solver(n-1,i);

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

    return 0;

}