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: 2.246604685154155x):
#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cmath>

using namespace std;



const int maxn = 1e5 + 5;

int dp[maxn][4];

int day[maxn][4];



int main(){

	int n; scanf("%d",&n);

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

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

			scanf("%d",&day[i][j]);

		}

	}



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

		dp[1][i] = day[1][i];

	}



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

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

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

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

	}



	int ans = max(dp[n][1], max(dp[n][2], dp[n][3]));



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



	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: 3.489321601490178x):
#include <bits/stdc++.h>

#define ll long long

#define ld long double

#define ull unsigned ll

#define ioi exit(0);

 

#define f first

#define s second

 

#define inf (int)1e9 + 7

 

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

 

#define mp(x,y) make_pair(x,y)

 

#define lb(x) lower_bound(x)

#define ub(x) upper_bound(x)

 

#define pb push_back

#define ppb pop_back

 

#define bitcoin __builtin_popcount

 

#define endl "\n"

 

#define in(x) insert(x)

 

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

 

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

 

#define pw2(x) (1ll<<x) //2^x

 

#define forit(it,v) for (typeof(v.begin()) it = v.begin(); it != v.end(); ++it)

 

#define sqr(x) ((x) * 1ll * (x))

 

#define UpdateRandom srand (time(NULL));

 

using namespace std;

 

const int N = (int)5e5 + 7, MOD = (int)1e9 + 7;



int n;

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

int dp[N][10];



int main(){

		

	NFS

	

	cin >> n;

	

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

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

	}

	

	for(int i=1;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])) << endl;



	ioi

}
// 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: 2.1591943115356607x):
#include <bits/stdc++.h>

#define ll long long

#define ld long double

#define ull unsigned ll

#define ioi exit(0);

 

#define f first

#define s second

 

#define inf (int)1e9 + 7

 

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

 

#define mp(x,y) make_pair(x,y)

 

#define lb(x) lower_bound(x)

#define ub(x) upper_bound(x)

 

#define pb push_back

#define ppb pop_back

 

#define bitcoin __builtin_popcount

 

#define endl "\n"

 

#define in(x) insert(x)

 

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

 

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

 

#define pw2(x) (1ll<<x) //2^x

 

#define forit(it,v) for (typeof(v.begin()) it = v.begin(); it != v.end(); ++it)

 

#define sqr(x) ((x) * 1ll * (x))

 

#define UpdateRandom srand (time(NULL));

 

using namespace std;

 

const int N = (int)5e5 + 7, MOD = (int)1e9 + 7;



int n;

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

int dp[N][10];



int main(){

		

	NFS

	

	cin >> n;

	

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

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

	}

	

	for(int i=1;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])) << endl;



	ioi

}
// 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: 2.2385348379334413x):
#include <bits/stdc++.h>

#define ll long long

#define ld long double

#define ull unsigned ll

#define ioi exit(0);

 

#define f first

#define s second

 

#define inf (int)1e9 + 7

 

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

 

#define mp(x,y) make_pair(x,y)

 

#define lb(x) lower_bound(x)

#define ub(x) upper_bound(x)

 

#define pb push_back

#define ppb pop_back

 

#define bitcoin __builtin_popcount

 

#define endl "\n"

 

#define in(x) insert(x)

 

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

 

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

 

#define pw2(x) (1ll<<x) //2^x

 

#define forit(it,v) for (typeof(v.begin()) it = v.begin(); it != v.end(); ++it)

 

#define sqr(x) ((x) * 1ll * (x))

 

#define UpdateRandom srand (time(NULL));

 

using namespace std;

 

const int N = (int)5e5 + 7, MOD = (int)1e9 + 7;



int n;

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

int dp[N][10];



int main(){

		

	NFS

	

	cin >> n;

	

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

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

	}

	

	for(int i=1;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])) << endl;



	ioi

}
// 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: 2.1887489716873754x):
#include <bits/stdc++.h>

#define ll long long

#define ld long double

#define ull unsigned ll

#define ioi exit(0);

 

#define f first

#define s second

 

#define inf (int)1e9 + 7

 

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

 

#define mp(x,y) make_pair(x,y)

 

#define lb(x) lower_bound(x)

#define ub(x) upper_bound(x)

 

#define pb push_back

#define ppb pop_back

 

#define bitcoin __builtin_popcount

 

#define endl "\n"

 

#define in(x) insert(x)

 

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

 

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

 

#define pw2(x) (1ll<<x) //2^x

 

#define forit(it,v) for (typeof(v.begin()) it = v.begin(); it != v.end(); ++it)

 

#define sqr(x) ((x) * 1ll * (x))

 

#define UpdateRandom srand (time(NULL));

 

using namespace std;

 

const int N = (int)5e5 + 7, MOD = (int)1e9 + 7;



int n;

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

int dp[N][10];



int main(){

		

	NFS

	

	cin >> n;

	

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

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

	}

	

	for(int i=1;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])) << endl;



	ioi

}
// 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: 3.531562322148979x):
/***************************************************************

Esse é um problema parecido com o problema da mochila.

Usando o código do exercício anterior

***************************************************************/

#include <bits/stdc++.h>



using namespace std;



int main(){

    int n=0, q=0;



    cin >> n;

    

    int matrix[n][3], resp[n][3];

    

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

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

            cin >> matrix[i][j];

            resp[i][j]=-1;

        }

    }



    resp[0][0]=matrix[0][0];

    resp[0][1]=matrix[0][1];

    resp[0][2]=matrix[0][2];

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

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

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

                    if (j==k){

                        continue;

                    }else{

                    

                    q=max(resp[i+1][k], resp[i][j]+matrix[i+1][k]);

                    }

                    // cout << "k: " << k << " j: " << j << endl;

                resp[i+1][k]=q;

                }

                // cout << "To no j" << endl;

        }

        // cout << "To no i" << endl;

    }

    

    

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

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

    //         cout << resp[i][j] << " ";  

    //     }

    //     cout << endl;

    // }



    //Não funciona, tem que imprimir o valor máximo na n-linha

    //cout << resp[n-1][2] << endl;

    int resposta=0;

    if(resp[n-1][0] >=resp[n-1][1] && resp[n-1][0] >=resp[n-1][2]){

        resposta = resp[n-1][0];

    }else if(resp[n-1][1] >=resp[n-1][0] && resp[n-1][1] >=resp[n-1][2]){

        resposta = resp[n-1][1];

    }else if(resp[n-1][2] >=resp[n-1][0] && resp[n-1][2] >=resp[n-1][1]){

        resposta = resp[n-1][2];

    }

    cout << resposta << endl;

    



}
// 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: 2.1853328949562028x):
/***************************************************************

Esse é um problema parecido com o problema da mochila.

Usando o código do exercício anterior

***************************************************************/

#include <bits/stdc++.h>



using namespace std;



int main(){

    int n=0, q=0;



    cin >> n;

    

    int matrix[n][3], resp[n][3];

    

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

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

            cin >> matrix[i][j];

            resp[i][j]=-1;

        }

    }



    resp[0][0]=matrix[0][0];

    resp[0][1]=matrix[0][1];

    resp[0][2]=matrix[0][2];

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

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

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

                    if (j==k){

                        continue;

                    }else{

                    

                    q=max(resp[i+1][k], resp[i][j]+matrix[i+1][k]);

                    }

                    // cout << "k: " << k << " j: " << j << endl;

                resp[i+1][k]=q;

                }

                // cout << "To no j" << endl;

        }

        // cout << "To no i" << endl;

    }

    

    

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

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

    //         cout << resp[i][j] << " ";  

    //     }

    //     cout << endl;

    // }



    //Não funciona, tem que imprimir o valor máximo na n-linha

    //cout << resp[n-1][2] << endl;

    int resposta=0;

    if(resp[n-1][0] >=resp[n-1][1] && resp[n-1][0] >=resp[n-1][2]){

        resposta = resp[n-1][0];

    }else if(resp[n-1][1] >=resp[n-1][0] && resp[n-1][1] >=resp[n-1][2]){

        resposta = resp[n-1][1];

    }else if(resp[n-1][2] >=resp[n-1][0] && resp[n-1][2] >=resp[n-1][1]){

        resposta = resp[n-1][2];

    }

    cout << resposta << endl;

    



}
// 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: 2.265633894877278x):
/***************************************************************

Esse é um problema parecido com o problema da mochila.

Usando o código do exercício anterior

***************************************************************/

#include <bits/stdc++.h>



using namespace std;



int main(){

    int n=0, q=0;



    cin >> n;

    

    int matrix[n][3], resp[n][3];

    

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

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

            cin >> matrix[i][j];

            resp[i][j]=-1;

        }

    }



    resp[0][0]=matrix[0][0];

    resp[0][1]=matrix[0][1];

    resp[0][2]=matrix[0][2];

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

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

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

                    if (j==k){

                        continue;

                    }else{

                    

                    q=max(resp[i+1][k], resp[i][j]+matrix[i+1][k]);

                    }

                    // cout << "k: " << k << " j: " << j << endl;

                resp[i+1][k]=q;

                }

                // cout << "To no j" << endl;

        }

        // cout << "To no i" << endl;

    }

    

    

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

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

    //         cout << resp[i][j] << " ";  

    //     }

    //     cout << endl;

    // }



    //Não funciona, tem que imprimir o valor máximo na n-linha

    //cout << resp[n-1][2] << endl;

    int resposta=0;

    if(resp[n-1][0] >=resp[n-1][1] && resp[n-1][0] >=resp[n-1][2]){

        resposta = resp[n-1][0];

    }else if(resp[n-1][1] >=resp[n-1][0] && resp[n-1][1] >=resp[n-1][2]){

        resposta = resp[n-1][1];

    }else if(resp[n-1][2] >=resp[n-1][0] && resp[n-1][2] >=resp[n-1][1]){

        resposta = resp[n-1][2];

    }

    cout << resposta << endl;

    



}
// 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: 2.215245335297368x):
/***************************************************************

Esse é um problema parecido com o problema da mochila.

Usando o código do exercício anterior

***************************************************************/

#include <bits/stdc++.h>



using namespace std;



int main(){

    int n=0, q=0;



    cin >> n;

    

    int matrix[n][3], resp[n][3];

    

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

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

            cin >> matrix[i][j];

            resp[i][j]=-1;

        }

    }



    resp[0][0]=matrix[0][0];

    resp[0][1]=matrix[0][1];

    resp[0][2]=matrix[0][2];

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

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

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

                    if (j==k){

                        continue;

                    }else{

                    

                    q=max(resp[i+1][k], resp[i][j]+matrix[i+1][k]);

                    }

                    // cout << "k: " << k << " j: " << j << endl;

                resp[i+1][k]=q;

                }

                // cout << "To no j" << endl;

        }

        // cout << "To no i" << endl;

    }

    

    

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

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

    //         cout << resp[i][j] << " ";  

    //     }

    //     cout << endl;

    // }



    //Não funciona, tem que imprimir o valor máximo na n-linha

    //cout << resp[n-1][2] << endl;

    int resposta=0;

    if(resp[n-1][0] >=resp[n-1][1] && resp[n-1][0] >=resp[n-1][2]){

        resposta = resp[n-1][0];

    }else if(resp[n-1][1] >=resp[n-1][0] && resp[n-1][1] >=resp[n-1][2]){

        resposta = resp[n-1][1];

    }else if(resp[n-1][2] >=resp[n-1][0] && resp[n-1][2] >=resp[n-1][1]){

        resposta = resp[n-1][2];

    }

    cout << resposta << endl;

    



}
// 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: 3.540894685977x):
#include<bits/stdc++.h>



#define pb push_back

#define se second

#define fi first



using namespace std;



long long n, a[(int)1e5 + 7][3], d[(int)1e5 + 7][3];



int main(){

    cin >> n;

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

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

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

    d[1][2] = a[1][2];

    d[1][3] = a[1][3];

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

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

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

                if(h != j)

                    d[i][j] = max(d[i][j], d[i - 1][h] + a[i][j]);

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

}