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

#include <algorithm>

#include <cstring>

const int maxn = 3e5 + 5;

using namespace std;



int N;

int a[maxn][4];

int dp[maxn][4];

int main()

{

	scanf("%d", &N);

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

	{

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

		{

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

		}

	}

	memset(dp, 0, sizeof dp);

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

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

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

	{

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

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

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

	}

	int ans = 0;

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

		ans = max(dp[N][i], ans);

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

	return 0;	

}

// Faster C++ solution for the same problem (measured speedup: 5.139947567184677x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

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

using namespace std;

#define db(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<'\n'

const int N = 1e5;

int A[N+7][5], mem[N+7][5];



int dp(int i, int j) {

	if(i==0) return 0;

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

	if(j==1) return mem[i][j] = A[i][j] + max(dp(i-1, 2), dp(i-1, 3));

	if(j==2) return mem[i][j] = A[i][j] + max(dp(i-1, 3), dp(i-1, 1));

	else return mem[i][j] = A[i][j] + max(dp(i-1, 1), dp(i-1, 2));

}

int main(){

	ios_base::sync_with_stdio(0);

	int a,b,n,T,cas=0;

	cin>>n;

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

		cin>>A[i][1]>>A[i][2]>>A[i][3];

	}

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

	int res = max(dp(n,1), max(dp(n,2), dp(n,3)));

	cout<<res<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.032658020044192x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

}
// 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: 2.2011598909579475x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

}
// 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: 2.2198791369685797x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

}
// 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: 2.2199055921923123x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

}
// 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: 2.2199033628351232x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

}
// 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: 2.2198791369685797x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

}
// 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: 2.220001771047993x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

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



using namespace std;

typedef long long ll;

typedef double dd;

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

#define endl "\n"

#define clr(n, r) memset(n,r,sizeof(n));



void fast() {

    cin.tie(0);

    cin.sync_with_stdio(0);

}

const int MAX=1e5+10;

int n,a[MAX],b[MAX],c[MAX],dp[MAX][3];

int ans(int i,int prev){

    if(i==n)return 0;

    int path1=0,path2=0,path3=0;

    int &ret=dp[i][prev];

    if(prev!=-1&&ret!=-1)return ret;

    if(prev!=0){

        path1=a[i]+ans(i+1,0);

    }

    if(prev!=1){

        path2=b[i]+ans(i+1,1);

    }

    if(prev!=2){

        path3=c[i]+ans(i+1,2);

    }

    return ret=max(path1,max(path2,path3));

}



int main() {

    fast();

    cin>>n;

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

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

    }

    clr(dp,-1);

    cout<<ans(0,-1);

}


// Faster C++ solution for the same problem (measured speedup: 2.230370461362611x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

}
// Slower C++ solution:
/*

** author:mahmoud magdy

*/

#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define endl "\n"



inline int D()

{

    int m ;

    cin>>m;

    return m ;

}

inline ll lD()

{

    ll m ;

    cin>>m;

    return m ;

}

inline double dD()

{

    double m;

    cin>>m;

    return m;

}

void fast()

{

    ios_base :: sync_with_stdio( 0 ) ;

    cin.tie( 0 ) ;

    cout.tie( 0 ) ;

}

int n;

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

ll dp[100005][5];

ll solve(int i,int prv){

 if(i==n)return 0;

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

 ll x=0;

 if(prv!=1){

 x=a[i]+solve(i+1,1);

 }

 ll y=0;

 if(prv!=2)

 y=b[i]+solve(i+1,2);

 ll z=0;

 if(prv!=3)

 z=c[i]+solve(i+1,3);

 return dp[i][prv]=max({x,y,z});

}

int main()

{

    memset(dp,-1,sizeof dp);

    fast();

    n=D();

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

        a[i]=D();b[i]=D();c[i]=D();

    }

    cout<<solve(0,10)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.130393203026971x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

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

#include <queue>

#include <string>

#include <map>

#include <stdio.h>

#include <iomanip>

#include <math.h>

#include <vector>

#include <stack>

#include <algorithm>

#include <string.h>

#include <set>

#include <cstdlib>

#include <iostream>

#include <bits/stdc++.h>

#define pii pair<int,int> 

#define ll long long int 

#define sl(n) scanf("%lld",&n)

#define s(n) scanf("%d",&n)

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

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

#define fnm(n,m) for(ll i=n;i<m+1;++i)

#define pb push_back

#define pf printf

#define F first

#define S second

#define srt(v) sort(v.begin(),v.end());

#define srtc(v,cmp) sort(v.begin(),v.end(),cmp);

#define srta(v,n) sort(v,v+n);

const ll LL_INF=0x3f3f3f3f3f3f;

const ll MOD=1073741824;





using namespace std;





int T;

int d,arr[100005][5];

int dp[100005][5];



int gen(int i,int prev){ // swim:0, bugs:1, homework:2

    if(i == d)return 0;

    int &ret = dp[i][prev];

    if(~ret)return ret;

    ret = gen(i+1,(prev+1)%3)+arr[i][(prev+1)%3];

    ret = max( ret , gen(i+1,(prev+2)%3)+arr[i][(prev+2)%3]);

    if(!i)ret = max( ret , gen(i+1,(prev)%3)+arr[i][(prev)%3]);

    return ret;

}



int main(){

    // pf("\t\t\t***************************************************************************\n\n");

    // pf("\t\t\t***************************************************************************\n\n");

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

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

    memset(dp,-1,sizeof dp);

    s(d);

    f(i,d){

        s(arr[i][0]);

        s(arr[i][1]);

        s(arr[i][2]);

    }

    pf("%d\n",gen(0,0));



    return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 3.0051680580789135x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

}
// Slower C++ solution:


#include <bits/stdc++.h>

using namespace  std;



int N;

const int SIZE = 100005;

int mem [SIZE][4];

int arr [SIZE][3];





int dp (int i, int activ) {

  

  if (i == N) return 0;

  



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

  

  int ret = -1; 



  if (activ != 0) {

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

  }



  if (activ != 1) {

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

    

  }



  if (activ != 2) {

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

    

  }



  return  mem[i][activ] = ret;

}



int main (){ 



  cin  >> N ;

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

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

  }



  memset(mem, -1, sizeof mem);

  





  cout << dp(0,3);



}

// Faster C++ solution for the same problem (measured speedup: 2.617019325339656x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

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



using namespace std;



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

int memo[100007][4];

int n; 



int solve(int p, int last) {

    if(p == n) return 0;

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

    if(st == -1) {

        st = max({

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

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

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

        });

    }

    return st;

}

int main() {

    memset(memo, -1, sizeof memo);

    cin >> n;

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

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

    }



    cout << solve(0, 0);

}

// Faster C++ solution for the same problem (measured speedup: 2.623439993425312x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

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

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

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

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

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

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

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

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

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

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





}

int n;

vector<vector<ll>>v;

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

ll solve(int i, int c)

{

	if (i == n)return 0;

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

	if (~ret)return ret;

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

	{

		if (k == c)continue;

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

	}

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		cin >> n;

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

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

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

	clr(mem, -1);

	ex(solve(0, 4));

}

// Faster C++ solution for the same problem (measured speedup: 5.133528151203745x):
#include <bits/stdc++.h>

using namespace std;



int n, a[100001][3], dp[100001][3];



int main(){

	scanf("%d", &n);

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

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

	}

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

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

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

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

	}

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

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

#include <algorithm>

#include <cstring>

const int maxn = 3e5 + 5;

using namespace std;



int N;

int a[maxn][4];

int dp[maxn][4];

int main()

{

	scanf("%d", &N);

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

	{

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

		{

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

		}

	}

	memset(dp, 0, sizeof dp);

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

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

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

	{

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

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

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

	}

	int ans = 0;

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

		ans = max(dp[N][i], ans);

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

	return 0;	

}

// Faster C++ solution for the same problem (measured speedup: 5.056515109722515x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

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

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

    ios_base::sync_with_stdio(0);

    cin.tie(NULL);

    long long a,fr,sc,th,bef1=0,bef2=0,bef3=0,ov1,ov2,ov3;

    cin>>a;

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

    {

        cin>>fr>>sc>>th;

        ov3=max(bef1,bef2)+th;

        ov2=max(bef1,bef3)+sc;

        ov1=max(bef3,bef2)+fr;

        bef1=ov1;

        bef2=ov2;

        bef3=ov3;

    }

    cout<<max(ov1,max(ov2,ov3));

    return 0;

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

using namespace std;

#define db(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<'\n'

const int N = 1e5;

int A[N+7][5], mem[N+7][5];



int dp(int i, int j) {

	if(i==0) return 0;

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

	if(j==1) return mem[i][j] = A[i][j] + max(dp(i-1, 2), dp(i-1, 3));

	if(j==2) return mem[i][j] = A[i][j] + max(dp(i-1, 3), dp(i-1, 1));

	else return mem[i][j] = A[i][j] + max(dp(i-1, 1), dp(i-1, 2));

}

int main(){

	ios_base::sync_with_stdio(0);

	int a,b,n,T,cas=0;

	cin>>n;

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

		cin>>A[i][1]>>A[i][2]>>A[i][3];

	}

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

	int res = max(dp(n,1), max(dp(n,2), dp(n,3)));

	cout<<res<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.983431426203039x):
#include <bits/stdc++.h>



using namespace std;



int main()

{

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

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

    ios_base::sync_with_stdio(0);

    cin.tie(NULL);

    long long a,fr,sc,th,bef1=0,bef2=0,bef3=0,ov1,ov2,ov3;

    cin>>a;

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

    {

        cin>>fr>>sc>>th;

        ov3=max(bef1,bef2)+th;

        ov2=max(bef1,bef3)+sc;

        ov1=max(bef3,bef2)+fr;

        bef1=ov1;

        bef2=ov2;

        bef3=ov3;

    }

    cout<<max(ov1,max(ov2,ov3));

    return 0;

}