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

#include <algorithm>

#include <iostream>

#include <iomanip>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <vector>

#include <string>

#include <climits>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <deque>

#include <stdio.h>

//#define int long long

using namespace std;

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

#define endl '\n'

#define OnlineJudgec

#define F first

#define S second

#define  precision(n)  cout << fixed << setprecision(n)

#define clr(v,d ) memset(v, d, sizeof(v))

typedef long long ll;

typedef vector<int> vi;

void init()

{

    cin.tie(0);

    cin.sync_with_stdio(0);

    cout.tie(0);

    #ifdef OnlineJudge

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

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

    #endif

}

ll S(ll x){ll s = 0;while(x){s += x % 10;x /= 10;}return s;} /// sum of digits.

ll powe(ll a , ll n){if(n == 1)return a;return a * powe(a ,n-1);}

ll com(int n,int m){ll z = 1;for(int i = 0; i < m;i++){z = z * ( n - i) / (i + 1) ;}return z;}

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

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

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

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

/// <<------------------------------------------------------------------------------>>

const int mx = 100009;

int n , k;

int arr[mx][4];

int dp[mx][4];

int solve(int idx , int last)

{

     if(idx == n)return 0;

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

     if(~ret) return ret;

     int A = 0 ,B = 0 ,C = 0;

     if(last != 0)

        A = solve(idx+1 , 0) + arr[idx][0];

     if(last != 1)

        B = solve(idx+1 , 1) + arr[idx][1];

     if(last != 2)

        C = solve(idx+1 , 2) + arr[idx][2];

     return ret = max(A,max(B,C));

}

/// <<------------------------------------------------------------------------------>>

int main()

{

    init();

    cin >> n;

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

    {

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

    }

    clr(dp,-1);

    cout<<solve(0,3);

}


// Faster C++ solution for the same problem (measured speedup: 13.631793691145502x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

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

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

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

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;

typedef long long ll;

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

#define x first

#define y second



vector<vector<int>> dp, arr;

int main() {

    fast;

    int n;

    cin>>n;

    arr.resize(n);

    dp.resize(n);

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

        arr[i].resize(3);

        dp[i].resize(3);

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

            cin>>arr[i][j];

            dp[i][j] = -1;

        }

    }

    for (int i = n - 1; i >= 0; i--) {

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

            if(i == n-1)

            {

                dp[i][j] = arr[i][j];

            } else{

                int temp, max = 0;

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

                    if(k == j)

                        continue;

                    temp = dp[i + 1][k];

                    if(temp > max)

                        max = temp;

                }

                dp[i][j] = max + arr[i][j];

            }

        }

    }

    sort(dp[0].begin(), dp[0].end());

    cout<<dp[0][2];

}

// Faster C++ solution for the same problem (measured speedup: 5.293433223908304x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

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

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

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

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define endl '\n'

#define mod 1000000007

#define maxN 1e5

const int inf=0x3f3f3f3f;



int main() {

	Omar45

	int n; cin >> n;

    int a[n+3], b[n+3], c[n+3], dp[n-1][3], res=0;

    for (int i=0; i<n; ++i) // 0-based //

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

    dp[0][0]=a[0]; dp[0][1]=b[0]; dp[0][2]=c[0];

    res= max(dp[0][0], max(dp[0][1], dp[0][2]));

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

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

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

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

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

            // last case //

            if (i==n-1) res = max(res, dp[i][j]);

        }

    }

    cout << res << endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.272051778424972x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

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

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

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

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

    {

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

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

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

    }

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



    return 0;

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

#include <algorithm>



#define ll long long



using namespace std;



int n;

ll arr[100005][3], dp[100005][3];



ll build()

{

	dp[0][0] = arr[0][0];

	dp[0][1] = arr[0][1];

	dp[0][2] = arr[0][2];

	ll ans = max(dp[0][0], max(dp[0][1], dp[0][2]));

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

	{

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

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

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

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

	}

	return ans;

}



int main()

{

	cin >> n;

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

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

			cin >> arr[i][j];

	cout << build() << '\n';

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.2200460666115704x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

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

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

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

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

    {

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

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

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

    }

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



    return 0;

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

#include <cmath>

#include <queue>

#include <set>

#include <string>

#include <vector>

#include <iomanip>

#include <algorithm>

#include <map>

#include <sstream>

using namespace std;

int dp[100000][5], a[100000], b[100000], c[100000];

int main()

{

    int n;

    cin >> n;

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

    {

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

    }

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

    dp[1][2] = b[1];

    dp[1][3] = c[1];

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

    {

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

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

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

    }

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.205636443284102x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

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

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

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

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

    {

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

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

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

    }

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



    return 0;

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

#define endl '\n'

using namespace std;

typedef long long ll;

const int c=1e5;

const ll mod=1e9+7;

/*bool p[1000];

void sieve(){

p[0]=p[1]=1;

for(int i=2;i<=100;i++){

    for(int j=i*i;j<=100;j+=i){

        p[j]=1;

    }

}

}*/

int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    int n;

    cin>>n;

    ll a[n+5]={},b[n+5]={},c[n+5]={};

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

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

    }

    ll dp[n][5]={};

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

    dp[0][2]=b[0];

    dp[0][3]=c[0];

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

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

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

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

    }

    cout<<max(dp[n-1][2],max(dp[n-1][1],dp[n-1][3]));

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.28318413058907x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

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

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

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

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

    {

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

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

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

    }

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



    return 0;

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



#define ll long long

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

#define pb push_back

const long double pi=3.141592654;

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

using namespace std;



const ll inf = 1e9 +5, mod = 1e9+7,N = 1e5+5;





int main()

{

	OSOS

	int n;

	cin>>n;

	int arr[N][3];

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

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

		{

			cin>>arr[i][ii];

		}





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

		{

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

			{

				int maxx = 0;

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

				{

					if(iii != ii)

					{

						maxx = max(arr[i-1][iii], maxx);

					}

				}

				arr[i][ii] += maxx;

			}

		}

		cout<<max(arr[n-1][0],max(arr[n-1][1],arr[n-1][2]));

}




// Faster C++ solution for the same problem (measured speedup: 5.389110887760284x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

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

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

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

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

    {

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

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

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

    }

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



    return 0;

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

#include <string>

#include <cmath> 

#include <algorithm>

using namespace std;

int main() {

	int n, a, b, c;

	cin >> n;

	int v[n + 1][3] = {0};

	cin >> a >> b >> c;

	v[1][0] = a;

	v[1][1] = b;

	v[1][2] = c;

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

		cin >> a >> b >> c;

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

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

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

	} 

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

	return 0; 

}

// Faster C++ solution for the same problem (measured speedup: 5.217694755972058x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

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

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

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

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;



int ar[100010][5],n;

int dp[100010][5];



int call(int i,int f)

{



    if(i==n)

        return 0;



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

        return dp[i][f];



    int ans=0;



    if(f==0)

        ans+=max(ar[i][1]+call(i+1,1),ar[i][2]+call(i+1,2));



    else if(f==1)

        ans+=max(ar[i][0]+call(i+1,0),ar[i][2]+call(i+1,2));

    else

        ans+=max(ar[i][1]+call(i+1,1),ar[i][0]+call(i+1,0));



        dp[i][f]=ans;

        return ans;



}



int main()

{



    cin>>n;



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

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



    memset(dp,-1,sizeof dp);



    int ans=max(max(call(0,0),call(0,1)),call(0,2));



    cout<<ans<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 2.9991711927733387x):
#include <bits/stdc++.h>



using namespace std;



const int maxn = 100001;



int value[maxn][3];

int dp[maxn][3];



int main() {

    int N;

    while (cin >> N) {

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

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

        }

        dp[0][0] = value[0][0];

        dp[0][1] = value[0][1];

        dp[0][2] = value[0][2];

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

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

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

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

        }

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

    }

    return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;



#define MAX 100000



int dp[MAX+5][4];

int val[MAX+5][4];

int n;



void initialize()

{

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

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

            dp[i][j] = -1;

}



int vacation(int i, int j)

{

    if(i<=0) return 0;



  /*  if(i==1)

    {

        return val[i][j];

    }

*/

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



    int mx = INT_MIN;



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

    {

        if(j==k) continue;



        mx = max(mx, vacation(i-1, k)+ val[i][j]);

    }





    dp[i][j] = mx;



    return mx;

}



int main()

{

    initialize();



    cin>>n;

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

    {

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

            cin>>val[i][j];

    }



    int retVal1 = vacation(n, 1);



    initialize();



    int retVal2 = vacation(n, 2);



    initialize();



    int retVal3 = vacation(n, 3);



    cout<<max({retVal1, retVal2, retVal3})<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.6258522378966793x):
#include <bits/stdc++.h>



using namespace std;



const int maxn = 100001;



int value[maxn][3];

int dp[maxn][3];



int main() {

    int N;

    while (cin >> N) {

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

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

        }

        dp[0][0] = value[0][0];

        dp[0][1] = value[0][1];

        dp[0][2] = value[0][2];

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

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

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

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

        }

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

    }

    return 0;

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

using namespace std;

int arr[100010][5];

int dp[100010][5];



int vac(int a, int b){

	if(a == 0) return 0;

	if(dp[a][b] != -1) return dp[a][b];

	

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

		if(i != b){

			dp[a][b] = max(dp[a][b], vac(a - 1,i) + arr[a][i]);

		}

	}

	

	return dp[a][b];

}



int main (){

	memset(dp,-1,sizeof dp);

	int n;

	cin >> n;

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

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

	}

	

	

	int ans = 0;

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

		ans = max(ans , vac(n - 1, i) + arr[n][i]);

	}

	

	cout << ans << endl;

	

}

// Faster C++ solution for the same problem (measured speedup: 3.001993037768972x):
#include <bits/stdc++.h>



using namespace std;



const int maxn = 100001;



int value[maxn][3];

int dp[maxn][3];



int main() {

    int N;

    while (cin >> N) {

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

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

        }

        dp[0][0] = value[0][0];

        dp[0][1] = value[0][1];

        dp[0][2] = value[0][2];

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

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

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

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

        }

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

    }

    return 0;

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

using namespace std;

#define INF INT_MAX

#define N 100005

#define K 3

int n, k;

int arr[N][K];

int dp[N][K];

int solve(int idx, int last_k){

    int res;

    if(idx == n)

        return 0;

    if(~last_k && ~dp[idx][last_k])

        return dp[idx][last_k];

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

        if(last_k != i){

            dp[idx][last_k] = max(dp[idx][last_k], arr[idx][i]+solve(idx+1, i));

        }

    }

    return dp[idx][last_k];

}

int main()

{

    cin >> n;

    memset(dp, -1, sizeof(dp[0][0]) * N * K);

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

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

            cin >> arr[i][j];

    cout << solve(0, -1);

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 2.193062275374062x):
#include <bits/stdc++.h>



using namespace std;



const int maxn = 100001;



int value[maxn][3];

int dp[maxn][3];



int main() {

    int N;

    while (cin >> N) {

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

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

        }

        dp[0][0] = value[0][0];

        dp[0][1] = value[0][1];

        dp[0][2] = value[0][2];

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

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

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

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

        }

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

    }

    return 0;

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

typedef long long ll;

using namespace std;

const int N = 2e5+10 , M = 5e5+10, OO = 0x3f3f3f3f ,  MOD = 1e9+7;

const double EPS = 1e-3;

#define M_PI 3.14159265358979323846

#define clr(arr,val) memset(arr , val , sizeof (arr))

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

#define rloop(i,n) for(int i = int(n)-1;i >= 0;i--)

#define range(i,a,b) for(int i = int(a);i <= int(b);i++)

#define pb push_back

#define F first

#define S second

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

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

typedef pair <int,int> pii ;

typedef vector < pii > vii ;

typedef vector < int > vi;

int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction

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

//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction

//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction

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

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







int n,x;

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

ll dp[N][4];





ll solve(int idx, int preC){ //preColour = (1 : a) -- (2 : b) -- (3 : c) -- (0 : invalid)



    if(idx == n) return 0;



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

    if(~ret) return ret;



    ret = -OO;

    if(preC != 1) ret = max(ret, solve(idx+1 , 1)+a[idx]);

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

    if(preC != 3) ret = max(ret, solve(idx+1 , 3)+c[idx]);



    return ret;



}







int main(){





    //ios_base::sync_with_stdio(0);

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





    scanf("%d",&n);

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

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



    memset(dp, -1 , sizeof dp);

    ll ans = solve(0 , 0);

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







    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.627882011008396x):
#include <bits/stdc++.h>



using namespace std;



const int maxn = 100001;



int value[maxn][3];

int dp[maxn][3];



int main() {

    int N;

    while (cin >> N) {

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

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

        }

        dp[0][0] = value[0][0];

        dp[0][1] = value[0][1];

        dp[0][2] = value[0][2];

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

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

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

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

        }

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

    }

    return 0;

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

#include <cstring>

using namespace std;

const int maxn=1e5+5;

const int inf=0x3f3f3f3f;

int dp[maxn][4];

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

int main()

{

   int n;

   cin>>n;

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

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

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

   }

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

   dp[1][2]=b[1];

   dp[1][3]=c[1];

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

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

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

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

   }

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

   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.6018032148809076x):
#include <bits/stdc++.h>



using namespace std;



const int maxn = 100001;



int value[maxn][3];

int dp[maxn][3];



int main() {

    int N;

    while (cin >> N) {

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

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

        }

        dp[0][0] = value[0][0];

        dp[0][1] = value[0][1];

        dp[0][2] = value[0][2];

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

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

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

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

        }

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

    }

    return 0;

}