stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



int n;

int a[N][3];

int mem[N][3];



int go(int idx , int lst){

    if(idx == n + 1)

        return 0;

    if(~mem[idx][lst])

        return mem[idx][lst];

    int best = 0;

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

        if(idx == 1 || lst != i){

            best = max(best , go(idx + 1 , i) + a[idx][i]);

        }

    }

    return mem[idx][lst] = best;

}



int main(){

    scanf("%d" , &n);

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

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

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

        }

    }

    memset(mem , -1 , sizeof mem);

    printf("%d\n" , go(1 , 0));

}


// Faster C++ solution for the same problem (measured speedup: 11.371741845966994x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

#include <algorithm>

#include <iostream>

#include <cstring>

#include <cstdlib>

#include <iomanip>

#include <vector>

#include <cstdio>

#include <stack>

#include <queue>

#include <deque>

#include <cmath>

#include <list>

#include <set>

#include <map>



using namespace std;



long long a[100010],b[100010],c[100010],f[100010][10];



int main(void)

{

    //freopen("test.in","r",stdin);

	//freopen("test.out","w",stdout);

	int n;

	cin >> n;

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

	{

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

	}

	f[1][1] = a[1];

	f[1][2] = b[1];

	f[1][3] = c[1];

	

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

	{

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

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

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

	}

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

	return 0;

}
























// Faster C++ solution for the same problem (measured speedup: 5.2109405298433105x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f3f3f3f

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define pb push_back

#define ms(a, x) memset(a, x, sizeof(a))

#define forn(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i) 

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int mod = 1e9;

const int N = 1e5+10;

/*=======================================================================*/



int n;

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

// ll ab, ac, ba, bc, ca, cb;

ll aa, bb, cc;

ll ans = 0;



int main() {

   scanf("%d", &n);

   for1(i, n) {

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

   }

   for1(i, n) {

      if(i == 1) {

         aa += a[i];

         bb += b[i];

         cc += c[i];

      } else {

         int ta = aa, tb = bb, tc = cc;

         aa = max(tb + a[i], tc + a[i]);

         bb = max(ta + b[i], tc + b[i]);

         cc = max(ta + c[i], tb + c[i]);

      }

   }

   ans = max(ans, aa);

   ans = max(ans, cc);

   ans = max(ans, bb);

   cout << ans << endl;

   return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.186205236425517x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	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;

//const int mod = 1e9+9;

/*ll gcd(ll x, ll y){return (y ? gcd(y, x%y) : x); }

*ll lcm(ll x, ll y){return x*(y/gcd(x,y));}

*void exEuclid(int a, int b){

	if(b==0){x = 1; y=0; d=a; return ;}

	exEuclid(b, a%b);

	int x1 = y;

	int y1 = x-(a/b)*y;

	x = x1; 100000

	y = y1;

} 

* */

//===========================================================//

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: 11.46845014692164x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

#define int long long

#define ll long long

#define ld long double

#define mod ((ll)1e9+7)

#define maxx ((ll)1e6+5)

#define pb push_back

#define ins insert

#define usefile freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);

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

#define test ll t; cin>>t; while(t--)

ll exp(ll x,ll y,ll p){ ll res=1; while(y) {if(y%2) res=(res*x%p)%p; x=(x*x)%p; y/=2; } return res;}

ll expo(ll x,ll y){ ll res=1; while(y) {if(y%2) res=(res*x%mod)%mod; x=(x*x)%mod; y/=2; } return res;}

/*int sv[1000000]={0};

ll fact[1000007]={0};

void facto() {fact[0]=1;fact[1]=1;for(int i=2;i<100007;i++)fact[i]=(fact[i-1]*i)%mod;}

void sve() {sv[0]=1;sv[1]=1;for(int i=2;i*i<=maxx;i++){if(!sv[i]){for(int j=2*i;j<=maxx;j+=i)sv[j]=1;}}}//0 hai to prime

ll ncr(ll n,ll r) {ll res=1; res=fact[n]; res=(res*(expo(fact[r],mod-2)))%mod; res=(res*(expo(fact[n-r],mod-2)))%mod; return res;}

ll npr(ll n,ll r) {ll res=1; res=fact[n]; res=(res*(expo(fact[n-r],mod-2)))%mod; return res;}*/

int d[100005][3];

using namespace std;

signed main()

{

    int n,k,i,j;

    cin >> n;

    int a[n][3];

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

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

    		cin >> a[i][j];

    d[0][0] = a[0][0];

    d[0][1] = a[0][1];

    d[0][2] = a[0][2];

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

    {

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

    	{

    		int p=0;

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

    		{

    			if(k!=j)

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

    		}

    		d[i][j] = p;

        }

   	}

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.199409817763926x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

     

#define ll long long

#define pb push_back

#define ff first

#define ss second

#define sz size()

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

#define allr(a) a.rbegin(), a.rend()

#define mod 1000000007

#define forn(i, n) for (int i = 0; i < (int) (n); ++i)

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

     

using namespace std;

     

int main(){

    IO;       

    int n;

    cin >> n;

    vector<array<int,3>> h(n);

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

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

            cin >> h[i][j];

        }

    }

 

    vector<array<ll,3>> dp(n);

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

        dp[0][j] = h[0][j];

    }

    

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

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

            dp[i][j] = 0;

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

                if (j == k) continue;

                dp[i][j] = max(dp[i][j], h[i][j] + dp[i - 1][k]);

            }

        }

    }

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

}

// Faster C++ solution for the same problem (measured speedup: 5.268688409702862x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

using namespace std;



int m;

const int mx = 1e5+5;

int tasks[mx][3];

int diario[3][mx];

int cosa(int a, int i){

	if(i == m){

     return 0;

     }



	if(diario[a][i] != -1) {

        return diario[a][i];

    }

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



        if(a==0){

		ans=max(ans, cosa(1, i+1)+tasks[i][0]);

		ans=max(ans, cosa(2, i+1)+tasks[i][0]);

	}

        if(a==1){

		ans=max(ans, cosa(0, i+1)+tasks[i][1]);

		ans=max(ans, cosa(2, i+1)+tasks[i][1]);

        }

	if(a==2){

		ans=max(ans, cosa(0, i+1)+tasks[i][2]);

		ans=max(ans, cosa(1, i+1)+tasks[i][2]);

	}



	return ans;

}

int main(){

	cin>>m;

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

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

	}

	memset(diario, -1, sizeof diario);

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 11.468586821215952x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

#include <cstdio>

#include <string>



using namespace std;





int n;

int dp[100007][4];

int a[100007][4];

int main() {

    

    scanf("%d",&n);

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

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

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

        }

    }

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

    dp[0][1] = a[0][1];

    dp[0][2] = a[0][2];

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

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

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

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

    }

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

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

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

}

// Faster C++ solution for the same problem (measured speedup: 5.1280270240177135x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	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 nn;

const int nxx = 1e5+5;

int acc[nxx][3];

int dpp[3][nxx];

int go(int aa, int ii)

{

	if(ii==nn) return 0;

	if(dpp[aa][ii] != -1) return dpp[aa][ii];

	int& ans = dpp[aa][ii];

	if(aa==0)

	{

		ans=max(ans, go(1, ii+1)+acc[ii][0]);

		ans=max(ans, go(2, ii+1)+acc[ii][0]);

	}

	if(aa==1)

	{

		ans=max(ans, go(0, ii+1)+acc[ii][1]);

		ans=max(ans, go(2, ii+1)+acc[ii][1]);

	}

	if(aa==2)

	{

		ans=max(ans, go(0, ii+1)+acc[ii][2]);

		ans=max(ans, go(1, ii+1)+acc[ii][2]);

	}



	return ans;



}

int main()

{

    cin>>nn;

    for(int ii=0; ii<nn; ii++) cin>>acc[ii][0]>>acc[ii][1]>>acc[ii][2];

    memset(dpp, -1, sizeof dpp);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 11.468575303799819x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

#include<algorithm>

using namespace std;

int main()

{

	int dp[2][3],a[3],n,i=0;

	cin >> n;

	dp[0][0] = 0;

	dp[0][1] = 0;

	dp[0][2] = 0;

	while (n--) {                                              

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

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

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

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

			i = 1 - i;

	}

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.190778253035739x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	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: 11.46845014692164x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

    #include <fstream>

    #include <iomanip>

    #include <iostream>

    #include <map>

    #include <math.h>

    #include <set>

    #include <stdio.h>

    #include <string>

    #include <utility>

    #include <vector>

     

    using namespace std;

     

    int main() {

      int n;

      cin >> n;

      vector<long long> act[3];

      act[0].resize(n);

      act[1].resize(n);

      act[2].resize(n);

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

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

     

      long long dp[n + 1][3];

      dp[0][0] = dp[0][1] = dp[0][2] = 0;

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

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

          dp[i][j] = 0;

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

            if (j == k)

              continue;

            dp[i][j] = max(dp[i][j], dp[i - 1][k] + act[j][i - 1]);

          }

        }

      }

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

      return 0;

    }

// Faster C++ solution for the same problem (measured speedup: 5.263036155263318x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

using namespace std;

int v[100009][3];

int dp[100009][3];

int n;

int vacation(int i, int j)

{

    if(i==n-1)

        return dp[i][j]=v[i][j];

    if(i>=n || j>=3)

        return 0;

    if(dp[i][j]!=-1)

        return dp[i][j];

    int temp = INT_MIN;

        if(j==0)

        {

             temp = max(vacation(i+1,j+1)+v[i][j],temp);

        temp = max(vacation(i+1,j+2)+v[i][j],temp);

 return dp[i][j]= temp;

        }

      else  if(j==1)

        {

             temp = max(vacation(i+1,0)+v[i][j],temp);

        temp = max(vacation(i+1,2)+v[i][j],temp);

 return dp[i][j]= temp;

        }

        else

        {

             temp = max(vacation(i+1,1)+v[i][j],temp);

        temp = max(vacation(i+1,0)+v[i][j],temp);

         return dp[i][j]= temp;

        }

}



int main()

{

    cin>>n;

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

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

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

    cout<<max(vacation(0,0),max(vacation(0,1),vacation(0,2)))<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 11.469083705209854x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[0]=in[0];dp[1]=in[1];dp[2]=in[2];

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.190809362967984x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

 }