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

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

#define ll long long

using namespace std;

const int N = 100010;

int dp[N][3];

int main(){

    fast;

	int n, i, a, b, c;

	cin >> n;

	memset(dp, 0, sizeof dp);

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

		cin >> a >> b >> c;

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

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

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

	}

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.1943257800866203x):
#pragma region

#include <algorithm>

#include <cstring>

#include <iostream>

#include <map>

#include <queue>

#include <stack>

#include <string>

#include <vector>

typedef long long ll;

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define rep(i, a, n) for (long long i = a; i <= n; ++i)

#define per(i, a, n) for (long long i = n; i >= a; --i)

#define IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)

template <typename T>

void read(T &x) {

    x = 0;

    char ch = getchar();

    ll f = 1;

    while (!isdigit(ch)) {

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

        ch = getchar();

    }

    while (isdigit(ch)) {

        x = x * 10 + ch - 48;

        ch = getchar();

    }

    x *= f;

}

using namespace std;

/*

 *                                                     __----~~~~~~~~~~~------___

 *                                    .  .   ~~//====......          __--~ ~~

 *                    -.            \_|//     |||\\  ~~~~~~::::... /~

 *                 ___-==_       _-~o~  \/    |||  \\            _/~~-

 *         __---~~~.==~||\=_    -_--~/_-~|-   |\\   \\        _/~

 *     _-~~     .=~    |  \\-_    '-~7  /-   /  ||    \      /

 *   .~       .~       |   \\ -_    /  /-   /   ||      \   /

 *  /  ____  /         |     \\ ~-_/  /|- _/   .||       \ /

 *  |~~    ~~|--~~~~--_ \     ~==-/   | \~--===~~        .\

 *           '         ~-|      /|    |-~\~~       __--~~

 *                       |-~~-_/ |    |   ~\_   _-~            /\

 *                            /  \     \__   \/~                \__

 *                        _--~ _/ | .-~~____--~-/                  ~~==.

 *                       ((->/~   '.|||' -_|    ~~-/ ,              . _||

 *                                  -_     ~\      ~~---l__i__i__i--~~_/

 *                                  _-~-__   ~)  \--______________--~~

 *                                //.-~~~-~_--~- |-------~~~~~~~~

 *                                       //.-~~~--\

 *                       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 */



#pragma endregion

const int maxn = 1e5 + 5;

int a[maxn], b[maxn], c[maxn], dp[maxn][3];

int main() {

    IO;

    int n;

    cin >> n;

    rep(i, 1, n) cin >> a[i] >> b[i] >> c[i];

    rep(i, 1, n) {

        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];

    }

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

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

using namespace std;



long long w1,w2,hap=0,maxi,aa,bb,cc,N,i,j,k,a[100005][5];

long long M[100005][5];





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cin>>N;

    for(i=0; i<100003; i++)

        for(j=0; j<4; j++)

            M[i][j]=0;



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

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

    M[N-1][0]=a[N-1][0];

    M[N-1][1]=a[N-1][1];

    M[N-1][2]=a[N-1][2];



    for(i=N-2; i>=0; i--)

    {

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

        {

            if(j==0)

            {

                w1=1;

                w2=2;

            }

            else if(j==1)

            {

                w1=0;

                w2=2;

            }

            else

            {

                w1=0;

                w2=1;

            }

        M[i][j]=a[i][j]+max(M[i+1][w1],M[i+1][w2]);

        }

    }

    aa=max(M[0][0],M[0][1]);

    aa=max(aa,M[0][2]);

    cout<<aa<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.119437428167746x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

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

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF][3];

long long mem[INF][3];



long long rec(int level,int pre)

{

    if(level==n) return 0;

    if(mem[level][pre]!=-1) return mem[level][pre];

    long long now=0;

    if(pre==0)

    {

        now=max(now,max(a[level][0]+rec(level+1,1),a[level][0]+rec(level+1,2)));

    }else

    if(pre==1)

    {

        now=max(now,max(a[level][1]+rec(level+1,0),a[level][1]+rec(level+1,2)));

    }else

    if(pre==2)

    {

        now=max(now,max(a[level][2]+rec(level+1,0),a[level][2]+rec(level+1,1)));

    }

    return mem[level][pre]=now;

}



int main()

{

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

    scanf("%d",&n);

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

    {

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

    }

    long long ans=max(rec(0,0),max(rec(0,1),rec(0,2)));

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

}


// Faster C++ solution for the same problem (measured speedup: 3.4343603326238545x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

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

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn = 1e5 + 5;



int a[maxn][4];

int dp[maxn][4];//d[i][j] i day choose j thing 's max happy



int main(){

	int n;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] ,dp[i - 1][3]) + a[i][1];

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

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

	}



	int ans = 0;

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

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

}

// Faster C++ solution for the same problem (measured speedup: 1.8107348210171195x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

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



using namespace std;



    const int n = 1e5+5;

    int num;

    int act[n][3];

    int dyn[3][n];

    

    int init(int a, int i){

        

    	if(i == num) return 0;

    	

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

    	int& res = dyn[a][i];

    	

    	if(a==0){

    	    

    		res=max(res, init(1, i+1)+act[i][0]);

    		res=max(res, init(2, i+1)+act[i][0]);

    	}

    	

    	if(a==1){

    	    

    		res=max(res, init(0, i+1)+act[i][1]);

    		res=max(res, init(2, i+1)+act[i][1]);

    	}

    	

    	if(a==2){

    	    

    		res=max(res, init(0, i+1)+act[i][2]);

    		res=max(res, init(1, i+1)+act[i][2]);

    	}

    	

    	

    	return res;

    }

    

    int main(){

        

    	cin>>num;

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

    	memset(dyn, -1, sizeof dyn);

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

    	return 0;

    	

    }

// Faster C++ solution for the same problem (measured speedup: 2.235388255503298x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

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

#include <string.h>

#define MAXN 100003



int array_a[MAXN][3];

int array_d[MAXN][3];



int max(int a,int b){

	return a>b? a:b;

}



int main(){

	memset(array_a,-1,sizeof(int)*MAXN*3);

	memset(array_d,-1,sizeof(int)*MAXN*3);

	int N;

	scanf("%d",&N);



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

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

	}



	array_d[0][0]=array_a[0][0];

	array_d[0][1]=array_a[0][1];

	array_d[0][2]=array_a[0][2];

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

		array_d[i][0]=array_a[i][0]+max(array_d[i-1][1],array_d[i-1][2]);

		array_d[i][1]=array_a[i][1]+max(array_d[i-1][0],array_d[i-1][2]);

		array_d[i][2]=array_a[i][2]+max(array_d[i-1][0],array_d[i-1][1]);

	}

	printf("%d",max(array_d[N-1][0],max(array_d[N-1][1],array_d[N-1][2])));

	return 0;



}

// Faster C++ solution for the same problem (measured speedup: 2.6236393640427425x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

}
// Slower C++ solution:
//       Author: Mostafa Mounir Shehab

 

#include <bits/stdc++.h>

using namespace std;

 

#define PI  3.14159265  // PI = acos(-1)

#define EPS (1e-10)

#define endl "\n"

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

typedef long long ll;

typedef long double ld;

//template <class Type>

 

 

void Fast(){

	std::ios_base::sync_with_stdio(0);

	cin.tie(NULL); cout.tie(NULL); cerr.tie(NULL);

}

 

inline int D(){

	int t;

	scanf("%d",&t);

	return t;

}

inline ll llD(){

	ll t;

	scanf("%lld",&t);

	return t;

}

 

 

ll MOD=1e9 + 7; 

const int N=1e5 + 5,M=50 + 1;

int arr[N][4];

ll dp[N][4];

ll OO=1e12; 

 

ll n,m,k,cnt,SU,idx,len,MN,MX;

ll l,r,mid;

ll a,b,c,d;



ll Solve(int idx,int row){

	

		

	if(idx == n)

		return 0;

	

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

	if(~ret)

		return ret;

	

	ret = -OO;

	

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

		if(row != i)

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

	

	return ret;

}







 

 

int main(){

	

	memset(dp, -1 ,sizeof dp);

	cin>>n;

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

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



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

	



	





}

// Faster C++ solution for the same problem (measured speedup: 4.2812200562507305x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

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

#include <bits/stdc++.h>

#include <unordered_map>

#define stop(x) {cout<<x<<endl;exit(0);}

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

#define szz(s) int(s.size())

#define endl '\n'

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

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

typedef long long ll;

using namespace std;



void file() {

    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#ifndef ONLINE_JUDGE

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

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

#endif

}





ll n;

ll mem[100004][4];

vector < vector < int > > v;



ll solve(int idx , int last) {

    if (idx == n)return 0;



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

    if (~ret)return ret;

    ret = 0;

    if (last != 0)ret = max(ret, solve(idx + 1, 0) + v[idx][0]);

    if (last != 1)ret = max(ret, solve(idx + 1, 1) + v[idx][1]);

    if (last != 2)ret = max(ret, solve(idx + 1, 2) + v[idx][2]);

    return ret;

}



int main() {

    //file();



    memset(mem, -1, sizeof mem);

    cin >> n;

    v = vector < vector < int > >(n, vector < int >(3));

    for (auto& e : v)

        for (auto& f : e)

            cin >> f;



    cout << solve(0,3) << endl;



}

// Faster C++ solution for the same problem (measured speedup: 4.290440679369422x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

}
// Slower C++ solution:
/******

author :- Mahmoud Tarek.

*******/



#include<bits/stdc++.h>



using namespace std ;



#define rg register

#define dbg(v) cerr << #v << " " << v ;

typedef long long ll ;



int n;

struct doW{

int w[5] ;

};



vector<doW> a ;

int mem[100002][6] ;



int solve(int i, int ch){

if(i == n) return 0 ;

int & ret = mem[i][ch] ;

if(~ret) return ret ;

ret = 0 ;

for(int j = 2 ; j <= 4 ; j++){

    if(ch != j){

        ret = max(ret, a[i].w[j] + solve(i + 1, j)) ;

    }

}

return ret ;

}



int main() {

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

    cin >> n ;

    a.resize(n) ;

    memset(mem, -1, sizeof mem) ;

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

        for(int j = 2 ; j <= 4 ; j++) cin >> a[i].w[j] ;

    }

    cout << solve(0, 1) ;



}



/*

  Empty set of troubles.

*/


// Faster C++ solution for the same problem (measured speedup: 3.446739319539101x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

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

using namespace std;



const int N=1e5+30;

typedef long long ll;

int n;

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

ll dp[N][5];

ll solve(int i,int last){

    if(i==n)return 0;

    int path1=-1 ,path2=-1,path3=-1;

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

    if(~ret)return ret;

    if(last!=1){

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

    }

    if(last!=2){

        path2=b[i]+solve(i+1,2);

    }

    if(last!=3){

        path3=c[i]+solve(i+1,3);

    }

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



}

int main()

{

    memset(dp,-1,sizeof dp);

    cin>>n;

    for(int i=0;i<n;i++)cin>>a[i]>>b[i]>>c[i];

    cout<<solve(0,4);



}




// Faster C++ solution for the same problem (measured speedup: 5.143210651752249x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

}
// Slower C++ solution:
/* MaTb3aa */

#include<bits/stdc++.h>

#define ll long long

using namespace std;

int n;

const int N = 100005;

vector<int>v;

ll dp[100005][4];

int A[N], B[N], C[N];

ll solve(int i,int lst) {

	if (i == n)return 0;

	ll& ret = dp[i][lst];

	if (~ret)return ret;

	if (lst != 1)ret = max(ret, solve(i + 1, 1) + A[i]);

	if (lst != 2)ret = max(ret, solve(i + 1, 2) + B[i]);

	if (lst != 3)ret = max(ret, solve(i + 1, 3) + C[i]);

	return ret;

}

int main() {

	cin >> n;

	for (int i = 0; i < n; i++)cin >> A[i] >> B[i] >> C[i];

	memset(dp, -1, sizeof dp);

	cout << solve(0, 0);

	

		 

}

// Faster C++ solution for the same problem (measured speedup: 4.267153423282619x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

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

using namespace std;

int n;

int arr[100005][3];



int dp[100005][4];



int solve(int i, int prev){

	if(i == n) return 0;

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



	dp[i][prev] = -1e7;

	int a = 0, b = 0, c = 0;

	if(prev != 0)

		a = solve(i+1, 0) + arr[i][0];

	if(prev != 1)

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

	if(prev != 2)

		c = solve(i+1, 2) + arr[i][2];



	dp[i][prev] = max(a, max(b, c));

	return dp[i][prev];

}



int main(){

	cin>>n;

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

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



	memset(dp, -1, sizeof dp);

	cout<<solve(0, 3)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6376639979904435x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

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

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

	cin.tie(NULL), cout.tie(NULL);

}

struct s {

	int x, y, z;

};



int n;

int v[(int)1e5][4];

int dp[(int)1e5][4];



int solve(int i = 0, int pre = 4) {

	if (i >= n)return 0;

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

	int ch1, ch2, ch3;

	ch1 = ch2 = ch3 = 0;

	if (pre != 1) {

		ch1 += solve(i + 1, 1) + v[i][1];

	}

	if (pre != 2) {

		ch2 += solve(i + 1, 2) + v[i][2];

	}

	if (pre != 3) {

		ch3 += solve(i + 1, 3) + v[i][3];

	}

	return dp[i][pre] = max({ ch1,ch2,ch3 });

}





int main()

{

	fast();



	cin >> n;

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

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

	}

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

	cout << solve();



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6495555161449946x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

	int n;

	scanf("%d",&n);

	int a[3][n];

	long long b[3][n];

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

	{

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

		{

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

		}

	}

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

	{

		b[i][0]=a[i][0];

	}

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

	{

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

		{

			b[t][i]=0;

		}

	}

	for(int i=1;i<n;i++)//从第二行开始遍历 

	{

		for(int t=0;t<3;t++)//遍历该行的每个状态 

		{

			for(int j=0;j<3;j++)//遍历上一行的每个状态 

			{

				if(j==t)continue;

				b[t][i]=max(b[t][i],b[j][i-1]+a[t][i]);

			}

		}

	}

	long long mm=b[0][n-1];

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

	{

		if(mm<b[i][n-1])mm=b[i][n-1];

	}

	printf("%lld",mm);

	return 0;

}