stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.1819324733016767x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

#define mx 100009

ll ar[mx][4];

ll br[mx][4];

ll n;



ll dp(){

    br[n][1]=ar[n][1];

    br[n][2]=ar[n][2];

    br[n][3]=ar[n][3];

    for(ll i=n-1;i>=1;i--){

        br[i][1]=ar[i][1]+max(br[i+1][2],br[i+1][3]);

        br[i][2]=ar[i][2]+max(br[i+1][1],br[i+1][3]);

        br[i][3]=ar[i][3]+max(br[i+1][1],br[i+1][2]);

    }

    return max(br[1][1],max(br[1][2],br[1][3]));

}



int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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



    //ll n;

    cin>>n;

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

        for(ll j=1;j<=3;j++){

            cin>>ar[i][j];

        }

    }

    cout<<dp()<<endl;

    return 0;

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

using namespace std;



int m;

const int mx = 1e5+5;

int tasks[mx][3];

int diario[3][mx];

int cosa(int a, int i){

	if(i == m){

     return 0;

     }



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

        return diario[a][i];

    }

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



        if(a==0){

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

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

	}

        if(a==1){

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

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

        }

	if(a==2){

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

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

	}



	return ans;

}

int main(){

	cin>>m;

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

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

	}

	memset(diario, -1, sizeof diario);

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.181958476299223x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

#define mx 100009

ll ar[mx][4];

ll br[mx][4];

ll n;



ll dp(){

    br[n][1]=ar[n][1];

    br[n][2]=ar[n][2];

    br[n][3]=ar[n][3];

    for(ll i=n-1;i>=1;i--){

        br[i][1]=ar[i][1]+max(br[i+1][2],br[i+1][3]);

        br[i][2]=ar[i][2]+max(br[i+1][1],br[i+1][3]);

        br[i][3]=ar[i][3]+max(br[i+1][1],br[i+1][2]);

    }

    return max(br[1][1],max(br[1][2],br[1][3]));

}



int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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



    //ll n;

    cin>>n;

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

        for(ll j=1;j<=3;j++){

            cin>>ar[i][j];

        }

    }

    cout<<dp()<<endl;

    return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;



int nn;

const int nxx = 1e5+5;

int acc[nxx][3];

int dpp[3][nxx];

int go(int aa, int ii)

{

	if(ii==nn) return 0;

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

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

	if(aa==0)

	{

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

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

	}

	if(aa==1)

	{

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

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

	}

	if(aa==2)

	{

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

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

	}



	return ans;



}

int main()

{

    cin>>nn;

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

    memset(dpp, -1, sizeof dpp);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.181956285050715x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

#define mx 100009

ll ar[mx][4];

ll br[mx][4];

ll n;



ll dp(){

    br[n][1]=ar[n][1];

    br[n][2]=ar[n][2];

    br[n][3]=ar[n][3];

    for(ll i=n-1;i>=1;i--){

        br[i][1]=ar[i][1]+max(br[i+1][2],br[i+1][3]);

        br[i][2]=ar[i][2]+max(br[i+1][1],br[i+1][3]);

        br[i][3]=ar[i][3]+max(br[i+1][1],br[i+1][2]);

    }

    return max(br[1][1],max(br[1][2],br[1][3]));

}



int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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



    //ll n;

    cin>>n;

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

        for(ll j=1;j<=3;j++){

            cin>>ar[i][j];

        }

    }

    cout<<dp()<<endl;

    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: 2.1819324733016767x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

#define mx 100009

ll ar[mx][4];

ll br[mx][4];

ll n;



ll dp(){

    br[n][1]=ar[n][1];

    br[n][2]=ar[n][2];

    br[n][3]=ar[n][3];

    for(ll i=n-1;i>=1;i--){

        br[i][1]=ar[i][1]+max(br[i+1][2],br[i+1][3]);

        br[i][2]=ar[i][2]+max(br[i+1][1],br[i+1][3]);

        br[i][3]=ar[i][3]+max(br[i+1][1],br[i+1][2]);

    }

    return max(br[1][1],max(br[1][2],br[1][3]));

}



int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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



    //ll n;

    cin>>n;

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

        for(ll j=1;j<=3;j++){

            cin>>ar[i][j];

        }

    }

    cout<<dp()<<endl;

    return 0;

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

using namespace std;

int v[100009][3];

int dp[100009][3];

int n;

int vacation(int i, int j)

{

    if(i==n-1)

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

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

        return 0;

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

        return dp[i][j];

    int temp = INT_MIN;

        if(j==0)

        {

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

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

 return dp[i][j]= temp;

        }

      else  if(j==1)

        {

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

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

 return dp[i][j]= temp;

        }

        else

        {

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

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

         return dp[i][j]= temp;

        }

}



int main()

{

    cin>>n;

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

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

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

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

}


// Faster C++ solution for the same problem (measured speedup: 2.1820530110713907x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

#define mx 100009

ll ar[mx][4];

ll br[mx][4];

ll n;



ll dp(){

    br[n][1]=ar[n][1];

    br[n][2]=ar[n][2];

    br[n][3]=ar[n][3];

    for(ll i=n-1;i>=1;i--){

        br[i][1]=ar[i][1]+max(br[i+1][2],br[i+1][3]);

        br[i][2]=ar[i][2]+max(br[i+1][1],br[i+1][3]);

        br[i][3]=ar[i][3]+max(br[i+1][1],br[i+1][2]);

    }

    return max(br[1][1],max(br[1][2],br[1][3]));

}



int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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



    //ll n;

    cin>>n;

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

        for(ll j=1;j<=3;j++){

            cin>>ar[i][j];

        }

    }

    cout<<dp()<<endl;

    return 0;

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

using namespace std;



const int MaxN = 100005;



int n;

int f[MaxN][3], val[MaxN][3];



void upd(int &a, int b) {

	if (b > a) a = b;

}



int main() {

	scanf("%d", &n);

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

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

	}

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

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

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

				if (j == k) continue;

				upd(f[i][j], f[i - 1][k] + val[i][j]);

			}

		}

	}

	int Ans = 0;

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

			upd(Ans, f[n][j]);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.188949466344356x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

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

using namespace std;

 

#define mp make_pair

#define pb push_back

#define st first

#define nd second

#define lb lower_bound

#define ub upper_bound

 

typedef long long ll;

typedef long double ld;

typedef pair<int,int> pi;

typedef pair<ll,ll> pll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<pi> vpi;

typedef vector<pll> vpll;

 

const ll oo=1e18;

const int MOD=1000000007;

const int INF=1000000000;

const ld EPS=0.000000001;

 

int dp[100005][3],n,a,b,c;

 

signed main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL);

    cout.tie(NULL);

    cin>>n;

    cin>>a>>b>>c;

    dp[1][0]=a;

    dp[1][1]=b;

    dp[1][2]=c;

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

        cin>>a>>b>>c;

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

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

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

    }

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.278231056110021x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

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

using namespace std;

int x[100000][3];

int a[100001][3], mx=0;

int calc(int pos, int u)

{

	if(pos==-1) return 0;

	if(a[pos][u]==-1)

	{

		//if(pos==0) a[pos][u]=mx(x[pos][(u+1)%3],x[pos][(u+2)%3]);

		a[pos][u]=max(calc(pos-1, (u+1)%3)+x[pos][(u+1)%3],calc(pos-1, (u+2)%3)+x[pos][(u+2)%3]);

	}

	return a[pos][u];

}



int main() {

	int n;

	cin>>n;

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

	{

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

		{

			cin>>x[i][u];

		}

	}

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

	{

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

		{

			a[i][u]=-1;

		}

	}

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

	{

		if(mx<calc(n, i) ) mx=calc(n-1, i);

	}

	cout<<mx;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.22444264825887x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

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

#define LL long long

using namespace std;

const int N =1e6;



LL int dp[N][4], a[N], b[N], c[N];



int main(){

    int n;

    cin >> n;

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

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

    }

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

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

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

    for(int i = 1; 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 << dp[i][0] << " " << dp[i][1] << " " << dp[i][2] << endl;

    }

    int ans = max(dp[n-1][0], max(dp[n-1][1], dp[n-1][2]));

    cout << ans << endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.234006853954947x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

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

using namespace std;



int main(){

    int n, i, j, k;

    scanf("%d",&n);

    int a[n+5], b[n+5], c[n+5];

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

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

    }

    int dp[n+5][3];

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

    for(i=1;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];

    }

    int ans  = 0;

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

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

    cout << ans << "\n";



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.215889412655526x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

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

using namespace std;



int main(){



    int n, flag;

    long long ans = 0;



    cin >> n;

    int a[n + 5], b[n + 5], c[n + 5], dp[n + 5][3];

    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];

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

    for(int i = 1; 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(dp[n][0], max(dp[n][1], dp[n][2])) << endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.226756210555024x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

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

using namespace std;



typedef long long  int ll;



#define s(t) scanf("%I64d",&t)

#define s2(t,t1) scanf("%I64d%I64d",&t,&t1)





ll n,a[100005][3];

int main()

{

    cin>>n;

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

    {

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

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

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

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



    }

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







}


// Faster C++ solution for the same problem (measured speedup: 5.21470029786272x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

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

#include <iostream>

#include <string>

#include <vector>

#include <algorithm>

#include <sstream>

#include <queue>

#include <deque>

#include <bitset>

#include <iterator>

#include <list>

#include <stack>

#include <map>

#include <set>

#include <functional>

#include <numeric>

#include <utility>



#include <cmath>

#include <iomanip>



using namespace std;



int main()

{

  ios::sync_with_stdio(false);

  cin.tie(0);

  int n;

  cin >> n;

  int dp[n][3]; // dp[i][j] represents the maximum happiness at the end of day i+1 if Taro did activity number j that day.

  int cur;

  cin >> cur;

  dp[0][0] = cur;

  cin >> cur;

  dp[0][1] = cur;

  cin >> cur;

  dp[0][2] = cur;

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

    cin >> cur;

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

    cin >> cur;

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

    cin >> cur;

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

  }

  cout << *max_element(dp[n-1], dp[n-1]+3) << "\n";

  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.284550533749261x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

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

using namespace std;

#define ff first

#define ss second

#define ll long long int

#define pii pair<int,int>

#define pll pair<ll,ll>

#define mem(x) memset(x, 0, sizeof x)

#define mem1(x) memset(x, -1, sizeof x)

#define MX 100005

ll arr[MX][3],arr2[MX],arr3[MX], dp[MX][3],n;

//ll rec(ll i, ll score)

//{

//    if(i==n-1) return score;

//    else if(i>=n) return 1e9;

//    else if(dp[i]<score && dp[i]!=-1) return dp[i];

//    else if(!i || dp[i]==-1 || score<dp[i])

//    {

//        ll ret1=1e9, ret2=1e9;

//        if(i+1<=n-1) ret1= rec(i+1, score+llabs(arr[i+1]-arr[i]));

//        if(i+2<=n-1) ret2= rec(i+2, score+llabs(arr[i+2]-arr[i]));

//        return dp[i]=min(ret1,ret2);

//    }

//}

int main()

{

    ll i,j,k,l, m;

    scanf("%lld",&n);

    for(i=0; i<n; i++) scanf("%lld%lld%lld",&arr[i][0],&arr[i][1],&arr[i][2]);

//    for(i=1; i<n; i++) dp[i]=1e9;

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

    {

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

        {

            ll mx=0;

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

                if(k!=j) mx=max(arr[i-1][k],mx);

            arr[i][j]+=mx;

        }

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.148437512264641x):
#include <cstdio>

#include <algorithm>



using namespace std;

const int maxn = 100000;

int arr[maxn][3], dp[maxn][3];



int main() {

    int n, a, b, c;

    scanf("%d", &n);

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

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

        arr[i][0] = a;

        arr[i][1] = b;

        arr[i][2] = c;

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

            if (i == 0) {

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

            } else {

                dp[i][j] = max(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) + arr[i][j];

            }

        }

    }

    int ans = -1;

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

        if (dp[n - 1][j] > ans) {

            ans = dp[n - 1][j];

        }

    }

    printf("%d", ans);

    return 0;

}