Probleme de informatica - enunturi si rezolvari

Hercule trebuie sa strabata un labirint cu capcane reprezentat de o matrice nXn. Pentru fiecare celula a labirintului, se cunoaste timpul in minute dupa care celula respectiva devine capcana. Dupa ce o celula devine capcana, Hercule moare daca intra in acea celula.
Hercule porneste din coltul stanga-sus al labirintului si trebuie sa ajunga in coltul dreapta jos. El are nevoie de un minut ca sa treaca dintr-o celula intr-una vecina si se poate deplasa in sus, in jos, spre stanga sau spre dreapta.
Sa se afiseze timpul minim in care poate Hercule sa strabata labirintul, numarul de drumuri de timp minim, precum si toate drumurile minime pe care le poate urma Hercule prin labirint de la intrare la iesire, astfel incat Hercule sa nu moara. Drumurile vor fi afisate ca matrici in care sunt indicati pasii lui Hercule.
Exemplu:
hercule.in
6 6
3 4 5 6 7 8
3 1 1 1 1 9
5 6 7 12 11 10
1 7 1 13 1 1
1 8 1 14 15 16
1 9 10 11 12 17
hercule.out
11 4
1 0 0 0 0 0
2 0 0 0 0 0
3 4 5 6 0 0
0 0 0 7 0 0
0 0 0 8 9 10
0 0 0 0 0 11

1 0 0 0 0 0
2 0 0 0 0 0
3 4 5 6 0 0
0 0 0 7 0 0
0 0 0 8 9 0
0 0 0 0 10 11

1 0 0 0 0 0
2 0 0 0 0 0
3 4 5 6 0 0
0 0 0 7 0 0
0 0 0 8 0 0
0 0 0 9 10 11

1 0 0 0 0 0
2 0 0 0 0 0
3 4 0 0 0 0
0 5 0 0 0 0
0 6 0 0 0 0
0 7 8 9 10 11

#include <fstream>
using namespace std;
ifstream is("hercule.in");
ofstream os("hercule.out");
const int di[]={-1,0,1,0};
const int dj[]={0,1,0,-1};
int a[100][100], x[100][100], m,n,pmin=10000,nr=0;

void citire()
{
    is>>m>>n;
    for(int i=1;i<=m;i++)
        for(int j=1;j<=n;j++)
            is>>a[i][j];
}

void afis(int x[100][100])
{
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=n;j++)
            os<<x[i][j]<<" ";
        os<<endl;
    }
    os<<endl;
}

void alege(int pas)//alege minimul si numarul de drumuri de lungime minima
{
    if(pas<pmin) { pmin=pas; nr=1;}
    else if(pas==pmin) nr++;
}

void back(int i, int j, int pas)
{
    x[i][j]=pas;
    if(i==m && j==n) alege(pas);
    else
    for(int d=0;d<4;d++)
    {
        int inou=i+di[d];
        int jnou=j+dj[d];
        if(pas<a[inou][jnou] && x[inou][jnou]==0 && pas<pmin) //inca nu e capcana, e pozitie noua si nu s-a depasit nr min de pasi
                back(inou,jnou,pas+1);
    }
    x[i][j]=0;
}

void backmin(int i, int j, int pas) //face doar drumurile de lungime minima
{
    x[i][j]=pas;
    if(i==m && j==n && pas==pmin) afis(x);
    else
    for(int d=0;d<4;d++)
    {
        int inou=i+di[d];
        int jnou=j+dj[d];
        if(pas<a[inou][jnou] && x[inou][jnou]==0 && pas<pmin)
                backmin(inou,jnou,pas+1);
    }
    x[i][j]=0;
}

int main()
{
    citire();
    back(1,1,1);
    os<<pmin<<" "<<nr<<endl;
    backmin(1,1,1);
    is.close();
    os.close();
    return 0;
}


sau

#include <fstream>
using namespace std;
ifstream is("hercule.in");
ofstream os("hercule.out");
const int di[]={-1,0,1,0};
const int dj[]={0,1,0,-1};
int a[100][100], x[100][100], m,n,pmin=10000,nr=0;

void citire()
{
    is>>m>>n;
    for(int i=1;i<=m;i++)
        for(int j=1;j<=n;j++)
            is>>a[i][j];
}

void afis(int x[100][100])
{
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=n;j++)
            os<<x[i][j]<<" ";
        os<<endl;
    }
    os<<endl;
}

void alege(int pas)
{
    if(pas<pmin) { pmin=pas; nr=1;}
    else if(pas==pmin) nr++;
}

void creste()
{
    for(int i=1;i<=m;i++)
        for(int j=1;j<=n;j++)
            a[i][j]++;
}

void scade()
{
    for(int i=1;i<=m;i++)
        for(int j=1;j<=n;j++)
            a[i][j]--;
}

void back(int i, int j, int pas)
{
    scade();
    x[i][j]=pas;
    if(i==m && j==n) alege(pas);
    else
    for(int d=0;d<4;d++)
    {
        int inou=i+di[d];
        int jnou=j+dj[d];
        if(a[inou][jnou]>0 && x[inou][jnou]==0 && pas<pmin)
            {
                back(inou,jnou,pas+1);
            }
    }
    x[i][j]=0;
    creste();
}

void backmin(int i, int j, int pas) //face doar drumurile de lungime minima
{
    scade();
    x[i][j]=pas;
    if(i==m && j==n && pas==pmin) afis(x);
    else
    for(int d=0;d<4;d++)
    {
        int inou=i+di[d];
        int jnou=j+dj[d];
        if(a[inou][jnou]>0 && x[inou][jnou]==0 && pas<pmin)
            {
                backmin(inou,jnou,pas+1);
            }
    }
    x[i][j]=0;
    creste();
}

int main()
{
    citire();
    back(1,1,1);
    os<<pmin<<" "<<nr<<endl;
    backmin(1,1,1);
    is.close();
    os.close();
    return 0;
}

29 ian 2025

Site-ul conține 884 de probleme rezolvate