Pagini recente » Cod sursa (job #523590) | Cod sursa (job #900758) | Cod sursa (job #2802075) | Cod sursa (job #1740215) | Cod sursa (job #342163)
Cod sursa(job #342163)
//Acest program rezolva problema romeo si julieta (Enunt in acelasi folder)
//ALGORITMUL LEE CU MATRICE (50 puncte pe infoarena)
//Depaseste timpul de executie (ineficient)
#include<fstream.h>
ifstream fin("rj.in");
ofstream fout("rj.out");
int n,m,x1,y1,x2,y2;
int a[101][101];
int rom[101][101];
int jul[101][101];
int l[8]={-1,-1,-1,0,1,1,1,0};
int c[8]={-1,0,1,1,1,0,-1,-1};
void citire()
{int i,j;
char c;
fin>>n>>m;
fin.get(c);
for(i=1;i<=n;i++)
{for(j=1;j<=m;j++)
{fin.get(c);
if(c==' ') a[i][j]=0;
else if(c=='X') a[i][j]=1;
else if(c=='R') {x1=i;
y1=j;
}
else {x2=i;
y2=j;
}
}
fin.get(c);
if(c!='\n')
while(c!='\n')
fin.get(c);
}
}
void lee()
{int i,j,ii,jj,k,pas=1;
typedef enum{DA,NU} boolean;
boolean gata=NU;
rom[x1][y1]=1;
jul[x2][y2]=1;
while(gata==NU)
{gata=DA;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
if(rom[i][j]==jul[i][j] && rom[i][j]==pas) {fout<<pas<<" "<<i<<" "<<j;
exit(0);
}
else if(rom[i][j]==pas) for(k=0;k<=7;k++)
{ii=i+l[k];
jj=j+c[k];
if(ii>=1 && ii<=n && jj>=1 && jj<=m)
if(a[ii][jj]==0 && rom[ii][jj]==0)
{rom[ii][jj]=pas+1;
gata=NU;
}
}
else if(jul[i][j]==pas) for(k=0;k<=7;k++)
{ii=i+l[k];
jj=j+c[k];
if(ii>=1 && ii<=n && jj>=1 && jj<=m)
if(a[ii][jj]==0 && jul[ii][jj]==0)
{jul[ii][jj]=pas+1;
gata=NU;
}
}
pas++;
}
}
int main()
{citire();
lee();
fin.close();
fout.close();
return 0;
}