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

#include <stdlib.h>

#include <string.h>

#include <math.h>

#include <queue>

#include <stack>

#include <vector>

#include <algorithm>

#include <iostream>



using namespace std;



const int MAXN = 1e5 + 5;

int n,m;

int a,b,c;

int d,e,f;

vector <int> A;

vector <int> B;

vector <int> C;

vector <int> dp[MAXN];



int main()

{

	scanf("%d",&n);

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

	{

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

		A.push_back(a);//第一天的a现在是a[0]

		B.push_back(b);

		C.push_back(c);

	}

	dp[0].push_back(A[0]);

	dp[0].push_back(B[0]);

	dp[0].push_back(C[0]);

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

	{

		d = max(dp[i - 1][1],dp[i - 1][2]);

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

		f = max(dp[i - 1][0],dp[i - 1][1]);

		dp[i].push_back(d + A[i]);

		dp[i].push_back(e + B[i]);

		dp[i].push_back(f + C[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];

	}

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

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.4783391456857675x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

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

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

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

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

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

    }

    ll n;

    cin >> n;

    ll arr[n+1][3];

    ll i,j;

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

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

            arr[i][j]=0;

        }

    }

    ll a,b,c;

    ll mx=-INF;

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

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

            cin >> arr[i][j];

            mx=max((arr[i-1][(j+1)%3]),(arr[i-1][(j+2)%3]));

            arr[i][j]+=mx;

        }

    }

    mx=max(arr[n][0],max(arr[n][1],arr[n][2]));

    cout << mx << endl;

    return 0;

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

using namespace std;



typedef signed long long ll;

const int MX = 1e5 + 5;

const int INF = 1e9 + 5;

int n, A[MX][3], dp[MX][4];



int solve(int pos, int prev) {

    if(pos >= n) return 0;

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

    int ans = 0;

    if(prev == 1) 

        ans = max(ans, max(A[pos][1] + solve(pos + 1, 2), A[pos][2] + solve(pos + 1, 3)));

    else if(prev == 2) 

        ans = max(ans, max(A[pos][0] + solve(pos + 1, 1), A[pos][2] + solve(pos + 1, 3)));

    else if(prev == 3) 

        ans = max(ans, max(A[pos][1] + solve(pos + 1, 2), A[pos][0] + solve(pos + 1, 1)));

    else

        ans = max(A[pos][0] + solve(pos + 1, 1), max(A[pos][1] + solve(pos + 1, 2), A[pos][2] + solve(pos + 1, 3)));

    return dp[pos][prev] = ans;

}



int main() {

    scanf("%d", &n);



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

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

    }

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

    int ans = solve(0, 0);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.557837795341488x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

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

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

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

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

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

    }

    ll n;

    cin >> n;

    ll arr[n+1][3];

    ll i,j;

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

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

            arr[i][j]=0;

        }

    }

    ll a,b,c;

    ll mx=-INF;

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

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

            cin >> arr[i][j];

            mx=max((arr[i-1][(j+1)%3]),(arr[i-1][(j+2)%3]));

            arr[i][j]+=mx;

        }

    }

    mx=max(arr[n][0],max(arr[n][1],arr[n][2]));

    cout << mx << endl;

    return 0;

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



using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;





int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i){

	if(i == n) return 0;

	if(dp[a][i] != -1) return dp[a][i];

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

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>n;

	for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.1824071435153427x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

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

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

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

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

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

    }

    ll n;

    cin >> n;

    ll arr[n+1][3];

    ll i,j;

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

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

            arr[i][j]=0;

        }

    }

    ll a,b,c;

    ll mx=-INF;

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

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

            cin >> arr[i][j];

            mx=max((arr[i-1][(j+1)%3]),(arr[i-1][(j+2)%3]));

            arr[i][j]+=mx;

        }

    }

    mx=max(arr[n][0],max(arr[n][1],arr[n][2]));

    cout << mx << endl;

    return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9; 



int m;

const int mx = 1e5+5;

int ac[mx][3];

int dp[3][mx];

int go(int a, int i){

	if(i == m) return 0;

	if(dp[a][i] != -1) return dp[a][i];

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

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>m;

	for(int i=0; i<m; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.18242918086192x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

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

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

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

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

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

    }

    ll n;

    cin >> n;

    ll arr[n+1][3];

    ll i,j;

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

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

            arr[i][j]=0;

        }

    }

    ll a,b,c;

    ll mx=-INF;

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

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

            cin >> arr[i][j];

            mx=max((arr[i-1][(j+1)%3]),(arr[i-1][(j+2)%3]));

            arr[i][j]+=mx;

        }

    }

    mx=max(arr[n][0],max(arr[n][1],arr[n][2]));

    cout << mx << endl;

    return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;



int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i)

{

	if(i==n) return 0;

	if(dp[a][i] != -1) return dp[a][i];

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

	if(a==0)

	{

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1)

	{

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2)

	{

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

	

}

int main()

{

    cin>>n;

    for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

    memset(dp, -1, sizeof dp);

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

}

	

// Faster C++ solution for the same problem (measured speedup: 2.1824071435153427x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

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

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

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

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

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

    }

    ll n;

    cin >> n;

    ll arr[n+1][3];

    ll i,j;

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

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

            arr[i][j]=0;

        }

    }

    ll a,b,c;

    ll mx=-INF;

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

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

            cin >> arr[i][j];

            mx=max((arr[i-1][(j+1)%3]),(arr[i-1][(j+2)%3]));

            arr[i][j]+=mx;

        }

    }

    mx=max(arr[n][0],max(arr[n][1],arr[n][2]));

    cout << mx << endl;

    return 0;

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

using namespace std;

const int mx=100010;

int arr[mx][3],n;

int dp[mx][3];

int f(int pos,int ty){

	if(pos==n)return 0;

	int &ans=dp[pos][ty];

	if(ans^-1)return ans;

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

		if(ty^i || !pos){

			ans=max(ans,f(pos+1,i)+arr[pos][i]);

		}

	}

	return ans;

}



int main(int argc, char const *argv[])

{

	memset(dp,-1,sizeof dp);

	cin>>n;

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

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

			cin>>arr[i][j];

		}

	}

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



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.183810882096874x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

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

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

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

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

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

    }

    ll n;

    cin >> n;

    ll arr[n+1][3];

    ll i,j;

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

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

            arr[i][j]=0;

        }

    }

    ll a,b,c;

    ll mx=-INF;

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

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

            cin >> arr[i][j];

            mx=max((arr[i-1][(j+1)%3]),(arr[i-1][(j+2)%3]));

            arr[i][j]+=mx;

        }

    }

    mx=max(arr[n][0],max(arr[n][1],arr[n][2]));

    cout << mx << endl;

    return 0;

}
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 2.183184084170851x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

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

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

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

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

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

    }

    ll n;

    cin >> n;

    ll arr[n+1][3];

    ll i,j;

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

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

            arr[i][j]=0;

        }

    }

    ll a,b,c;

    ll mx=-INF;

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

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

            cin >> arr[i][j];

            mx=max((arr[i-1][(j+1)%3]),(arr[i-1][(j+2)%3]));

            arr[i][j]+=mx;

        }

    }

    mx=max(arr[n][0],max(arr[n][1],arr[n][2]));

    cout << mx << endl;

    return 0;

}
// 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: 2.558094767615812x):
#include <bits/stdc++.h>

#include <algorithm>

#include <climits>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <iostream>

#include <math.h>

using namespace __gnu_pbds;

using namespace std;

#define ll long long

typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define vll vector<ll> 

#define nl cout << endl

#define fr(a,b) for(auto i=a;i<b;i++)

#define pll pair<ll,ll>

#define f first

#define s second

#define vpll vector<pll>

#define ump unordered_map

#define YES cout << "YES" << endl

#define NO cout << "NO" << endl

#define IN(a,b) ((a).find(b)!=(a).end())

#define pb push_back

#define mkp make_pair

#define PI M_PI

#define INF 1e18

#define deb(x) if(ON) cout << #x <<" : " << x << endl

#define fast ios_base::sync_with_stdio(false)

#define io cin.tie(NULL),cout.tie(NULL)

#define ON !false

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

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

const ll mod=1e9+7;

ll i,j;

ll add(ll a,ll b){

    return  ((a%mod)+(b%mod)+mod)%mod;

}

ll mul(ll a,ll b){

    return ((a%mod)*(b%mod)+mod)%mod;

}

void pause(ll id){

    cout << "Waiting " << id << endl;

    cin >> id;

}



const int mxn=1e5+10;

ll n,t;





signed main(){

    fast;

    io;

    if(!true){

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

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

    }

    ll n;

    cin >> n;

    ll arr[n+1][3];

    ll i,j;

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

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

            arr[i][j]=0;

        }

    }

    ll a,b,c;

    ll mx=-INF;

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

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

            cin >> arr[i][j];

            mx=max((arr[i-1][(j+1)%3]),(arr[i-1][(j+2)%3]));

            arr[i][j]+=mx;

        }

    }

    mx=max(arr[n][0],max(arr[n][1],arr[n][2]));

    cout << mx << endl;

    return 0;

}