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



using namespace std;



vector<vector<int> > v;

int ans[100005][3];

queue<pair<int,pair<int,int> > >s;

int maior,a=0;



void dp(int i, int bloq, int sum) {

  if(ans[i-1][bloq]>sum)

    return;

  

  if (i>=v.size())

    maior = max(sum, maior);

  else {

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

      if (bloq != j) {



        if(ans[i][j]==-1) {

          ans[i][j] = sum+v[i][j];

          s.push({sum+v[i][j],{i+1,j}});

        }

        else if (ans[i][j]<sum+v[i][j]) {

          ans[i][j] = sum+v[i][j];

          s.push({sum+v[i][j],{i+1,j}});

        }



        //dp(i+1, j, sum+v[i][j]);

      }

    }

  }

}



int main () {

  int n;

  cin>>n;

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

    int a,b,c;

    cin>>a>>b>>c;

    v.push_back({a,b,c});

  }



  maior=0;

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

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

    s.push({v[0][i],{1,i}});

    //dp(1, i, v[0][i]);

  }



  while(!s.empty()) {

    pair<int,pair<int,int> > aux = s.front();

    dp(aux.second.first, aux.second.second, aux.first);

    s.pop();

  }



  cout<<maior<<endl;



}




// Faster C++ solution for the same problem (measured speedup: 2.2373339097789233x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

int n,a[1000010],b[1000010], c[1000010],dp[1000010][5];

int main()

{

    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][3],max(dp[n][1],dp[n][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: 3.4923239136385003x):
#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;

}
// 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: 2.1610521441038437x):
#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;

}
// 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: 2.2404609373607625x):
#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;

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



using namespace std;



vector<vector<int> > v;

int ans[100005][3];

queue<pair<int,pair<int,int> > >s;

int maior,a=0;



void dp(int i, int bloq, int sum) {

  if(ans[i-1][bloq]>sum)

    return;

  

  if (i>=v.size())

    maior = max(sum, maior);

  else {

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

      if (bloq != j) {



        if(ans[i][j]==-1) {

          ans[i][j] = sum+v[i][j];

          s.push({sum+v[i][j],{i+1,j}});

        }

        else if (ans[i][j]<sum+v[i][j]) {

          ans[i][j] = sum+v[i][j];

          s.push({sum+v[i][j],{i+1,j}});

        }



        //dp(i+1, j, sum+v[i][j]);

      }

    }

  }

}



int main () {

  int n;

  cin>>n;

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

    int a,b,c;

    cin>>a>>b>>c;

    v.push_back({a,b,c});

  }



  maior=0;

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

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

    s.push({v[0][i],{1,i}});

    //dp(1, i, v[0][i]);

  }



  while(!s.empty()) {

    pair<int,pair<int,int> > aux = s.front();

    dp(aux.second.first, aux.second.second, aux.first);

    s.pop();

  }



  cout<<maior<<endl;



}




// Faster C++ solution for the same problem (measured speedup: 2.190632233930821x):
#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;

}
// 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: 1.6019974382570126x):
#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;

}
// 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: 3.496530818659843x):
#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;

}
// 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: 2.163655379468384x):
#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;

}
// 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: 2.243159829731741x):
#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;

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



using namespace std;



vector<vector<int> > v;

int ans[100005][3];

queue<pair<int,pair<int,int> > >s;

int maior,a=0;



void dp(int i, int bloq, int sum) {

  if(ans[i-1][bloq]>sum)

    return;

  

  if (i>=v.size())

    maior = max(sum, maior);

  else {

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

      if (bloq != j) {



        if(ans[i][j]==-1) {

          ans[i][j] = sum+v[i][j];

          s.push({sum+v[i][j],{i+1,j}});

        }

        else if (ans[i][j]<sum+v[i][j]) {

          ans[i][j] = sum+v[i][j];

          s.push({sum+v[i][j],{i+1,j}});

        }



        //dp(i+1, j, sum+v[i][j]);

      }

    }

  }

}



int main () {

  int n;

  cin>>n;

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

    int a,b,c;

    cin>>a>>b>>c;

    v.push_back({a,b,c});

  }



  maior=0;

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

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

    s.push({v[0][i],{1,i}});

    //dp(1, i, v[0][i]);

  }



  while(!s.empty()) {

    pair<int,pair<int,int> > aux = s.front();

    dp(aux.second.first, aux.second.second, aux.first);

    s.pop();

  }



  cout<<maior<<endl;



}




// Faster C++ solution for the same problem (measured speedup: 2.1932711019089166x):
#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;

}
// 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: 2.1826069974644278x):
#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;

}
// 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: 3.5391528577654277x):
# 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] } );

}
// 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: 2.1900299229737823x):
# 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] } );

}
// 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: 2.270503517215539x):
# 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] } );

}