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

#define _CRT_SECURE_NO_WARNINGS

#include <bits/stdc++.h>

#include <unordered_set>

#include <unordered_map>



using namespace std;



typedef long long ll;

typedef long long ull;

typedef pair<ll, ll> ii;



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

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

#define endl		"\n"

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

#define mk			make_pair	



void File()

{

#ifndef ONLINE_JUDGE

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

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

#endif

#ifdef ONLINE_JUDGE

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

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

#endif

}

void fast()

{

	ios::sync_with_stdio(NULL);

	cout.tie(NULL);

	cin.tie(NULL);

	//File();

}

const double pi = 2 * acos(0.0);

const ll oo = 0x3f3f3f3f;

const int MOD = 1e9 + 7;

const int nn = 1e5 + 15;

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

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



int n, k;

vector<pair<int, pair<int, int>>>v;

ll dp[nn][4];

ll solve(int idx, int last)

{

	if (idx >= n)

		return 0;

	ll &ret = dp[idx][last];

	if (~ret)

		return ret;

	ret = 0;

	if (last != 1)

		ret = max(ret, solve(idx + 1, 1) + v[idx].first);



	if (last != 2)

		ret = max(ret, solve(idx + 1, 2) + v[idx].second.first);



	if (last != 3)

		ret = max(ret, solve(idx + 1, 3) + v[idx].second.second);



	return ret;

}

int main()

{

	fast();

	cin >> n;

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

	{

		int x, y, z;

		cin >> x >> y >> z;

		v.push_back(mk(x, mk(y, z)));

	}

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

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.6356460664323074x):
#include <iostream>

#include <cstring>

#include <stdio.h>

using namespace std;

const int N=1e+5;

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

int memo[N][4];

int dp(int i,int t)

{

    if(i==n)

        return 0;

    int &ret=memo[i][t];

    if(~ret)

        return ret;

    ret=0;

    if(t!=0)

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

    if(t!=1)

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

    if(t!=2)

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

    return ret;





}

int main()

{

    memset(memo,-1,sizeof memo);

    scanf("%d",&n);

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

    {

         scanf("%d %d %d",a+i,b+i,c+i);

    }

    printf("%d\n",dp(0,3));

    return 0;

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

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

#define fast ios_base::sync_with_stdio(false); cin.tie(NULL)

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

struct edge

{

    ll  a,b,c;

};

vector<edge> v;

ll  dp[100009][4];

ll  call(ll  ind, ll  pre)

{

    if(ind==v.size()) return 0;

    ll  &ret=dp[ind][pre];

    if(ret!=-1) return ret;

    ret=0;

    ll  a=0,b=0,c=0;

    auto x=v[ind];

        if(pre==1)

        {

            a=x.b+call(ind+1,2);

            b=x.c+call(ind+1,3);

        }

        else if(pre==2)

        {

            a=x.a+call(ind+1,1);

            b=x.c+call(ind+1,3);

        }

        else if(pre==3)

        {

            a=x.a+call(ind+1,1);

            b=x.b+call(ind+1,2);

        }

        else

        {

            a=x.a+call(ind+1,1);

            b=x.b+call(ind+1,2);

            c=x.c+call(ind+1,3);

        }

    return ret=max(a,max(b,c));

}

int  main()

{

    ll  n ;

    cin>>n;

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

    {

        ll  a,b,c; cin>>a>>b>>c;

        v.pb({a,b,c});

    }

    memset(dp,-1,sizeof dp);

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







}


// Faster C++ solution for the same problem (measured speedup: 1.6263383495645645x):
#include <iostream>

#include <cstring>

#include <stdio.h>

using namespace std;

const int N=1e+5;

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

int memo[N][4];

int dp(int i,int t)

{

    if(i==n)

        return 0;

    int &ret=memo[i][t];

    if(~ret)

        return ret;

    ret=0;

    if(t!=0)

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

    if(t!=1)

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

    if(t!=2)

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

    return ret;





}

int main()

{

    memset(memo,-1,sizeof memo);

    scanf("%d",&n);

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

    {

         scanf("%d %d %d",a+i,b+i,c+i);

    }

    printf("%d\n",dp(0,3));

    return 0;

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

#define ll long long

using namespace std;



ll n;

std::vector<ll> ara[100010];

ll dp[100010][4];



ll fnc(ll idx,ll last){

  if(idx == n) return 0;

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



  ll res1 = - (1<<25);

  ll res2 = - (1<<25);

  ll res3 = - (1<<25);



  if(last == 0){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 1){

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 2){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res3 = max(res3,ara[idx][2] + fnc(idx+1,3));

  }else if(last == 3){

    res1 = max(res1,ara[idx][0] + fnc(idx+1,1));

    res2 = max(res2,ara[idx][1] + fnc(idx+1,2));

  }

  dp[idx][last] = max(res1, max(res2,res3));

  return dp[idx][last];

}



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

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

  cin >> n;

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

    int a,b,c;

    cin >> a >> b >> c;

    ara[i].push_back(a);

    ara[i].push_back(b);

    ara[i].push_back(c);

  }



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

  return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.76190420223894x):
#include <iostream>

#include <cstring>

#include <stdio.h>

using namespace std;

const int N=1e+5;

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

int memo[N][4];

int dp(int i,int t)

{

    if(i==n)

        return 0;

    int &ret=memo[i][t];

    if(~ret)

        return ret;

    ret=0;

    if(t!=0)

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

    if(t!=1)

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

    if(t!=2)

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

    return ret;





}

int main()

{

    memset(memo,-1,sizeof memo);

    scanf("%d",&n);

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

    {

         scanf("%d %d %d",a+i,b+i,c+i);

    }

    printf("%d\n",dp(0,3));

    return 0;

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



using namespace std;

typedef long long int lli;



lli n, a[100005], b[100005], c[100005];



lli dp[100005][4];



lli solve(lli i, lli last){

    if(i == n) return dp[i][last] = 0;

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

    lli ans1, ans2, ans3;

    ans1 = ans2 = ans3 = LONG_LONG_MIN;

    if(last != 1) ans1 = a[i] + solve(i+1, 1);

    if(last != 2) ans2 = b[i] + solve(i+1, 2);

    if(last != 3) ans3 = c[i] + solve(i+1, 3);



    return dp[i][last] = max(ans1, max(ans2, ans3));

}





int main(){

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

    int T=1, cs=1;

    //scanf("%d", &T);

    while(T--){

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

        scanf("%lld", &n);

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

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

        }

        printf("%lld\n", solve(0, 0));



    }



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.6196098018194376x):
#include <iostream>

#include <cstring>

#include <stdio.h>

using namespace std;

const int N=1e+5;

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

int memo[N][4];

int dp(int i,int t)

{

    if(i==n)

        return 0;

    int &ret=memo[i][t];

    if(~ret)

        return ret;

    ret=0;

    if(t!=0)

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

    if(t!=1)

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

    if(t!=2)

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

    return ret;





}

int main()

{

    memset(memo,-1,sizeof memo);

    scanf("%d",&n);

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

    {

         scanf("%d %d %d",a+i,b+i,c+i);

    }

    printf("%d\n",dp(0,3));

    return 0;

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

#include <cmath>

#include <cstring>

using namespace std;

const int maxn=1e5+1;

int n;

int a[maxn][4];

int f[maxn][4];



int main(){

	cin>>n;

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

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

			cin>>a[i][j];

		}

	}

	f[0][1]=a[0][1];f[0][2]=a[0][2];f[0][3]=a[0][3];

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

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

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

				if (k==j) continue;

				f[i][k]=max(f[i][k],f[i-1][j]+a[i][k]);

			}

		}

	}

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

//		for(int j=1..3) {

//			

//			for(int k=1..3) {

//				if (k==j) continue;

//				dp[i][j]=min(dp[i][j], dp[i-1][j]+a[i][k]);

//			} 

//		} 

	

//	}

	int max=-1;

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

		if (f[n-1][i]>max) max=f[n-1][i];

	}

	cout<<max<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.496904897246061x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include <limits.h>

#include <algorithm>  

#include <numeric>

 

#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;

using namespace std;

 

#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

#define endl '\n'

using ll = long long;

 

const int precision = 3;

const int modulo =  1000000007; // 10^9 + 7

const ll lmodulo =  1000000007; // 10^9 + 7

 

const double EPS = 1e-9;

const ll inf = 1e18;



void solve () {

	int n; cin >> n;

	vector<vector<int>> days(n, vector<int>(3));

	for (auto &it : days) {

		cin >> it[0] >> it[1] >> it[2];

	}



	vector<vector<int>> dp(n, vector<int>(3));

	dp[0] = days[0];



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

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

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

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

	}



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

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

	cout << ret << endl;

}



int main() {

  ios_base::sync_with_stdio(false);

  cin.tie(nullptr);

  cout.tie(nullptr);

  cout.precision(precision);

  cout.setf(ios_base::fixed);

  solve();

  return 0;

}




// Faster C++ solution for the same problem (measured speedup: 4.596058339587623x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

using namespace std;



int n;

int dp[100100][4];

vector<int> arr[100100];



int main() {

    cin >> n;

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

        int k;

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

            cin >> k;

            arr[i].push_back(k);

        }

    }



    for (int i = 0; i < 3; ++i) dp[1][i] = arr[1][i];



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

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



            if ( j == 0 ) 

                dp[pos][j] = arr[pos][j] + max(dp[pos - 1][1], dp[pos - 1][2]);

            

            else if ( j == 1 )

                dp[pos][j] = arr[pos][j] + max(dp[pos - 1][0], dp[pos - 1][2]);



            else

                dp[pos][j] = arr[pos][j] + max(dp[pos - 1][0], dp[pos - 1][1]);

        }

    }



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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 15.887892292917238x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  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: 4.49189204321285x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include<stdio.h>

#include<vector>

#pragma GCC optimize(2)

typedef long long ll;

typedef unsigned long long ull;

const int mod=1000;

const int maxn=1e5+5;

const ll INF=100000000000005;

using namespace std;

ll n,m,p;

ll qpow(ll a,ll b){ll ans=1;while(b){if(b&1) ans=(ans*a)%mod;b/=2;a=(a*a)%mod;}return ans;}

struct node{

    ll a,b,c;

}save[maxn];

ll dp[maxn][4];

int main()

{

    scanf("%lld",&n);

    for(int i=1;i<=n;i++) scanf("%lld%lld%lld",&save[i].a,&save[i].b,&save[i].c);

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

    {

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

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

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

    }

    ll res=0;

    for(int i=1;i<=3;i++) res=max(dp[n][i],res);

    printf("%lld\n",res);

    return 0;

}

/***

10 4

40 10 20 70 80 10 20 70 80 60

***/


// Faster C++ solution for the same problem (measured speedup: 4.432781115610787x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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



using namespace std;

int main(){

	long long n,ans=0;

	scanf("%lld", &n);

	long long a[n+5],b[n+5],c[n+5],check[n+5][4];

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

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

	memset(check,0,sizeof(check));

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

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

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

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

	}

	ans=max(check[n][1], check[n][2]);

	ans=max(ans,check[n][3]);

	printf("%lld", ans);

} 

// Faster C++ solution for the same problem (measured speedup: 4.434969328296066x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include<map>

#include<cmath>

#include<ctime>

#include<queue>

#include<stack>

#include<cstdio>

#include<string>

#include<vector>

#include<random>

#include<cstdlib>

#include<cstring>

#include<iomanip>

#include<iostream>

#include<algorithm>

#include<unordered_map>

#define endl "\n"

#define fi first

#define se second

#define db double

#define gcd __gcd

#define pb push_back

#define mp make_pair

#define lowbit(x) (x & (-x))

#define PII  pair<int, int> 

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

#define debug(x) cout << #x << " = " << x << endl

#define rep(i, a, b) for(__typeof(b) i = a; i <= (b); i++)

#define Rep(i, a, b) for(__typeof(a) i = a; i >= (b); i--)

#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

template<class T> inline T qmin(T a, T b) { return a < b ? a : b; }

template<class T> inline T qmax(T a, T b) { return a > b ? a : b; }

typedef long long ll;

typedef unsigned long long ull;

const db eps = 1e-9;

const db PI = acos(-1);

const int inf = 0x3f3f3f3f;

const int mod = (int)1e9 + 7;

const int maxn = (int)1e5 + 5;//remember to modify it, No RE or MLE

const ll INF = 0x3f3f3f3f3f3f3f3f;

using namespace std;

ll qpow(ll x, ll y) { ll ret = 1; for( ; y; y >>= 1, x = (x * x) % mod) if(y & 1) ret = (ret * x) % mod; return ret; }



inline int read(){

	char ch = getchar();

	int x = 0, f = 1;

	while(ch < '0' || ch > '9'){

		if(ch == '-') f = -1;

		ch = getchar();

	}

	while('0' <= ch && ch <= '9'){

		x = x * 10 + ch - '0';

		ch = getchar();

	}

	return x * f;

}



int n;

int val[3][maxn];

int dp[maxn][3];



int main()

{

	n = read();

	rep(i, 1, n) rep(j, 0, 2) val[j][i] = read();

	rep(i, 1, n){

		rep(j, 0, 2){

			rep(k, 0, 2){

				if(k == j) continue;

				dp[i][j] = qmax(dp[i][j], dp[i-1][k] + val[j][i]);

			}

		}

	}

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.41496813335663x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

using namespace std;



#define MAX_N 100005

int N, h[MAX_N][3], memo[MAX_N][3];



int main() {

  scanf("%i", &N);

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

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

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

	}

  }

  memset(memo, -1, sizeof memo);

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

  	memo[0][c] = h[0][c];

  }

  for (int d = 1; d < N; d++) {

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

  	  int &opt = memo[d][c];

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

  	    if (cc != c) {

  	      opt = max(opt, memo[d - 1][cc] + h[d][c]);

		  /*

  	      if (d == 1) {

  	        printf("%i %i %i %i\n", memo[d - 1][cc], h[d][c], cc, c);

		  }

		  */

		}

	  }

	}

  }

  /*

  puts("\ndebug");

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

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

  	  printf("%i ", memo[d][c]);

	}

	puts("");

  }

  puts("end-debug\n");

  */

  int ans = -1;

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

  	ans = max(ans, memo[N - 1][c]);

  }

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

  

}

// Faster C++ solution for the same problem (measured speedup: 9.831437333408184x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

int main()

{

    int n,i,j,k;

    int a[100005][5];

    scanf("%d",&n);

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

    scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

    int dp[100005][5];

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

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

    dp[1][2] = a[1][2];

    dp[1][3] = a[1][3];

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

    {

        for(j = 1;j < 4;j++)

        {

            for(k = 1;k < 4;k++)

            {

                if(k != j)

                {

                    if(dp[i][j] < dp[i - 1][k] + a[i][j])

                    dp[i][j] = dp[i - 1][k] + a[i][j];

                }

            }

        }

    }

    printf("%d\n",max(dp[n][1],max(dp[n][2],dp[n][3])));

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 10.19269773925124x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

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

#include<vector>

using namespace std;

int main(){

  int n;

  cin>>n;

  vector<vector<int>>task;

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

    task.push_back({0});

  

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

    int a;

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

      cin>>a;

      if(j==0)task[0].push_back(max(task[1][i],task[2][i])+a);

      if(j==1)task[1].push_back(max(task[0][i],task[2][i])+a);

      if(j==2)task[2].push_back(max(task[1][i],task[0][i])+a);

    }

  }

  cout<< max(task[0][task[0].size()-1],max(task[1][task[1].size()-1],task[2][task[2].size()-1]))<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 4.551495582550525x):
#include<cstdio>

#include<algorithm>

int main(){

  int j=1,x,da[3][100001],fun[3][2];

  scanf("%d",&x);

  fun[0][0] = fun[1][0] = fun[2][0] = 0;

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

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

    fun[0][j] = std::max(fun[1][j^1],fun[2][j^1]) + da[0][i];

    fun[1][j] = std::max(fun[0][j^1],fun[2][j^1]) + da[1][i];

    fun[2][j] = std::max(fun[1][j^1],fun[0][j^1]) + da[2][i];

    j^=1;

  }

  j^=1;

  printf("%d\n",std::max(std::max(fun[0][j],fun[1][j]),fun[2][j]));

  

  return 0;

}