stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 9.244287619053333x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

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

using namespace std;

typedef long long LL;



int a[100005],b[100005],c[100005];

int dp[100007][5];



int Solve(int idx, int n, int prev){

	

	if(idx == n)return 0;

	

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

	

	int ans = -1;

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

		if(i == prev)continue;

		if(i == 1)ans = max(ans, a[idx] + Solve(idx+1, n, i));

		else if(i == 2)ans = max(ans, b[idx] + Solve(idx+1, n, i));

		else ans = max(ans, c[idx] + Solve(idx+1, n, i));		

	}

	return dp[idx][prev] = ans;

}



int main(){

	int n; cin>>n;

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

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

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

	}

	int ret = -1;

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

		ret = max(ret, Solve(0,n,i));

	}

	cout<<ret<<endl;



//	cout<<Solve(0,n,0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.967900316875567x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}
// 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: 3.4755326272199905x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}
// 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: 2.5557739888222204x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}
// 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: 2.180646255432961x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}
// 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: 2.180668274998575x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}
// 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: 2.180646255432961x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}
// 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: 2.182048861399734x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}
// 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: 2.1814225692093165x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}
// 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: 2.5560307537569487x):
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

using ii = pair<int, int>;

using iii = pair<ii, int>;

#define vt vector

#define pq priority_queue

#define pb push_back

#define pf push_front

#define mp make_pair

#define fi first

#define se second

#define __lcm(a, b) a * b / __gcd(a, b)

const int MOD = 1e9 + 7;

const int INF = 2e9;

const int N = 1e5 + 5;



int n;

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

int dp[N][5];



void getDP() {

    // i = index, j = now

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

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

            int &sol = dp[i][j];

            if(j == 0) 

                sol = max(dp[i - 1][1] + a[i - 1],

                      max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]));

            else if(j == 1) 

                sol = max(dp[i - 1][2] + b[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else if(j == 2) 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][3] + c[i - 1]);

            else 

                sol = max(dp[i - 1][1] + a[i - 1],

                          dp[i - 1][2] + b[i - 1]);

        }

    }

}



void solve() {

    cin >> n;

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

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

    }

    getDP();

    cout << max(dp[n][1] + a[n], max(dp[n][2] + b[n], dp[n][3] + c[n])) << "\n";

}



int main() {

 

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    

    solve();

    

}