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

using namespace std;

#define ll long long

#define ull unsigned long long

#define mst(a, b) memset((a), (b), sizeof(a))

#define mp(a, b) make_pair(a, b)

#define pi acos(-1)

#define pii pair<int, int>

#define pb push_back

const int INF = 0x3f3f3f3f;

const double eps = 1e-10;

const int maxn = 1e5 + 107;

const int maxm = 5e5 + 10;

int s[maxn][3];

int dp[maxn][3];

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

    int n;

    cin >> n;

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

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

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

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

    {

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

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

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

    }

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.275301320104644x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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;

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: 11.500056436978234x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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;



const int nmax =100010;



int act[3];

int dp[nmax][3];



int main()

{

    ios_base::sync_with_stdio(0);

    cin.tie(0);



    int n;cin>>n;

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

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

    {

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

        dp[i][0]=max(b + act[0], c + act[0]);

        dp[i][1]=max(a + act[1], c + act[1]);

        dp[i][2]=max(a + act[2], b + act[2]);

        a = dp[i][0];

        b = dp[i][1];

        c = dp[i][2];

    }

    int sol = max({ dp[n][0],dp[n][1],dp[n][2] });

    cout<<sol;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.268954259900224x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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 dp[100007][5], a[100007][5];



int main() {

	int n;

	cin >> n;

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

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

	}  

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

	}

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

}

// Faster C++ solution for the same problem (measured speedup: 5.205500212127673x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

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

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

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

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

#define pb push_back

#define eb emplace_back

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

#define F first

#define S second

#define endl '\n'

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

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



const int N=1e5+5;

int a[N][3];

int dp[N][4];

int n;

int dfs(int x,int la){

	if(x==n+1) return 0;

	if(dp[x][la]) return dp[x][la];

	int res=0;

	forn(i,3){

		if(i==la) continue;

		res=max(res,a[x][i]+dfs(x+1,i));

	}

	return dp[x][la]=res;

}

int main() {

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	cin>>n;

	for1(i,n){

		forn(j,3) cin>>a[i][j];

	}

	cout<<dfs(1,4);

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.313155531538678x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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 <bits/stdc++.h>



using namespace std;

int a[100005][5], p[100005][5];



int main()

{

    int n, i, j;

    cin>>n;

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

    {

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

        {

            cin>>a[i][j];

        }

    }

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

    {

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

        {

            if(i==0)

            {

                p[i][j]=a[i][j];

            }

            else if(j==0)

            {

                p[i][j]=a[i][j]+max(p[i-1][1],p[i-1][2]);

            }

            else if(j==1)

            {

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

            }

            else if(j==2)

            {

                p[i][j]=a[i][j]+max(p[i-1][1],p[i-1][0]);

            }

        }

    }

    n--;

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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.215782227017543x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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 all(v) v.begin(), v.end()

#define mp make_pair

using namespace std;



using ll = long long;

using ii = pair<int, int>;

const int INF = (int)10e9;

int dp[3][100005];



int main() {

    std::ios::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);



    int N;

    cin>>N;

    vector<int> a(N), b(N), c(N);

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

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

    }



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

    dp[1][0] = b[0];

    dp[2][0] = c[0];

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

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

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

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

    }

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



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.325712816725573x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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;

const int oo = 1e9;

int number;

const int nx = 1e5+5;

int ac[nx][3];

int dyn[3][nx];

int init(int a, int i){

	if(i == number) 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)+ac[i][0]);

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

	}

	if(a==1){

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

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

	}

	if(a==2){

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

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

	}

	

	return res;

}

int main(){

	cin>>number;

	for(int i=0; i<number; i++) cin>>ac[i][0]>>ac[i][1]>>ac[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: 11.500181938780491x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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 main(){  

    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<long long,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: 5.261170875723272x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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;

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 x, int i){

	if(i==n) return 0;

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

	int& ans = dp[x][i];

	if(x==0)

	{

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(x==1)

	{

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(x==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: 11.500056436978234x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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;



ll arr[100009][3];

int dp[100009][2][2][2];



long long VAC1(int n,int a, int b, int c)

{



    if(!n && a) return arr[n][0];

    else if(!n && b) return arr[n][1];

    else if(!n && c) return arr[n][2];

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



    else

    {



        ll x,y,z;

        if(a)

        {

            x = VAC1(n-1,0,1,0)+arr[n][0];

            y = VAC1(n-1,0,0,1)+arr[n][0];



        }



        else if(b)

        {

            x = VAC1(n-1,1,0,0)+arr[n][1];

            y = VAC1(n-1,0,0,1)+arr[n][1];

        }



        else if(c)

        {

            x = VAC1(n-1,1,0,0)+arr[n][2];

            y = VAC1(n-1,0,1,0)+arr[n][2];

        }



        dp[n][a][b][c] = max(x,y);;

        return dp[n][a][b][c];





    }

}



int main()

{

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

    int n;

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

    cin >> n;

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

    {

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

        {

            cin >> arr[i][j];

        }

    }

    ll res = VAC1(n-1,1,0,0);

    res= max(res, VAC1(n-1,0,1,0));

    res = max(res,VAC1(n-1,0,0,1));



    cout << res << endl;

}


// Faster C++ solution for the same problem (measured speedup: 21.983263329331255x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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 ll long long

ll a[100009],b[100009],c[100009];

int dp[100009][3];

int main() {

    ios::sync_with_stdio(false);

    int n;cin>>n;

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

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

    memset(dp,0x3f, sizeof(dp));

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

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

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

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

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

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 11.566114064655894x):
#include <cstdio>

#include <algorithm>

using std::max;

const int maxn = 100003;

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

int main() {

	int i,n;

    scanf("%d", &n);

    for(int 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(int 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 n;

int dp[100100][4];

vector<int> arr[100100];



int main() {

    cin >> n;

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

        int k;

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

            cin >> k;

            arr[i].push_back(k);

        }

    }



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



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

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



            if ( j == 0 ) 

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

            

            else if ( j == 1 )

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



            else

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

        }

    }



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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.5342120565423567x):
#include<iostream>

#include<algorithm>

using namespace std;

int dp[100010][3];

int a[100010][3];

int main()

{

    int n;

    cin >> n;

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

    {

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

    }

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

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

    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][1], dp[i-1][0]);

    }

    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_N 100005

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



int main() {

  scanf("%i", &N);

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

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

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

	}

  }

  memset(memo, -1, sizeof memo);

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

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

  }

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

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

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

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

  	    if (cc != c) {

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

		  /*

  	      if (d == 1) {

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

		  }

		  */

		}

	  }

	}

  }

  /*

  puts("\ndebug");

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

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

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

	}

	puts("");

  }

  puts("end-debug\n");

  */

  int ans = -1;

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

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

  }

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

  

}

// Faster C++ solution for the same problem (measured speedup: 2.186972552196974x):
#include<iostream>

#include<algorithm>

using namespace std;

int dp[100010][3];

int a[100010][3];

int main()

{

    int n;

    cin >> n;

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

    {

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

    }

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

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

    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][1], dp[i-1][0]);

    }

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

    return 0;

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

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

int main()

{

    int n,i,j,k;

    int a[100005][5];

    scanf("%d",&n);

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

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

    int dp[100005][5];

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

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

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

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

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

    {

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

        {

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

            {

                if(k != j)

                {

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

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

                }

            }

        }

    }

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.267333802030666x):
#include<iostream>

#include<algorithm>

using namespace std;

int dp[100010][3];

int a[100010][3];

int main()

{

    int n;

    cin >> n;

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

    {

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

    }

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

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

    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][1], dp[i-1][0]);

    }

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

    return 0;

}