Just an advice, especially when you want someone else to read your code, try to make your variable names meaningful and put comments that explain what every part is supposed to do, I guarantee you will get faster response and will even encourage you to debug your code yourself :)
On Wed, Mar 13, 2013 at 11:10 AM, anup1pma <[email protected]> wrote: > hi > http://www.lightoj.com/volume_showproblem.php?problem=1434 > m solved this problem using DFS but getting TLE > here is my source code > > #include<iostream> > #include<vector> > #include<string> > #include<cstring> > #include<cstdio> > using namespace std; > > int indexx[26][1802]; > int arr[20];//0-DL:1-D:2-DR:3-L:4-R:5-UL:6-U:7-UR > int i,r,c; > char b[40][40]; > int vt[40][40]; > char ss[40]; > > int dfs(int rr,int cc,int idx) > { > int n=strlen(ss); > if(rr>0&&cc>0&&ss[idx+1]==b[rr-1][cc-1]) > { > > > if((idx+1==n-1)||dfs(rr-1,cc-1,idx+1)>=0) > {arr[idx]=0;return 50;} > > > } > if(rr>0&&ss[idx+1]==b[rr-1][cc]) > { > if((idx+1==n-1)||dfs(rr-1,cc,idx+1)>=0) > {arr[idx]=1;return 50;} > > } > if(rr>0&&cc<c-1&&ss[idx+1]==b[rr-1][cc+1]) > { > if((idx+1==n-1)||dfs(rr-1,cc+1,idx+1)>=0) > {arr[idx]=2;return 50;} > > } > if(cc>0&&ss[idx+1]==b[rr][cc-1]) > { > if((idx+1==n-1)||dfs(rr,cc-1,idx+1)>=0) > {arr[idx]=3;return 50;} > > } > if(cc<c-1&&ss[idx+1]==b[rr][cc+1]) > { > if((idx+1==n-1)||dfs(rr,cc+1,idx+1)>=0) > {arr[idx]=4;return 50;} > > } > if(rr<r-1&&cc>0&&ss[idx+1]==b[rr+1][cc-1]) > { > if((idx+1==n-1)||dfs(rr+1,cc-1,idx+1)>=0) > {arr[idx]=5;return 50;} > > } > if(rr<r-1&&ss[idx+1]==b[rr+1][cc]) > { > if((idx+1==n-1)||dfs(rr+1,cc,idx+1)>=0) > {arr[idx]=6;return 50;} > > } > if(rr<r-1&&cc<c-1&&ss[idx+1]==b[rr+1][cc+1]) > { > if((idx+1==n-1)||dfs(rr+1,cc+1,idx+1)>=0) > {arr[idx]=7;return 50;} > > } > if(ss[idx+1]==b[rr][cc]) > { > if((idx+1==n-1)||dfs(rr,cc,idx+1)>=0) > {arr[idx]=8;return 50;} > > } > return -6; > > } > > > int main() > { > int cases,cs,N,i,j,pp; > int ft,ac,ar,fff,k; > char ch; > //string name; > scanf("%d",&cases); > for(int I=0;I<cases;I++) > { > scanf("%d %d",&r,&c); > getchar(); > for(i=0;i<r;i++) > { > scanf("%s",b[i]); > } > memset(indexx,-1,sizeof(indexx)); > for(i=0;i<r;i++){ > for(j=0;j<c;j++) > { > k=0; > while(indexx[(int)(b[i][j]-'A')][k]>=0) > k+=2; > indexx[(int)(b[i][j]-'A')][k]=i; > indexx[(int)(b[i][j]-'A')][k+1]=j; > > //cout<<indexx[b[i][j]-'A'][k]<<" "; > } > // cout<<endl; > } > > > > scanf("%d",&N); > getchar(); > printf("Case %d:\n",I+1); > > > //cin.get(); > while(N--) > { > > pp=0; > scanf("%s",ss); > ch=ss[0]; > ar=-1; > ac=-1; > fff=1; > k=0; > while(indexx[(int)(ch-'A')][k]>=0) > { > i=indexx[(int)(ch-'A')][k]; > j=indexx[(int)(ch-'A')][k+1]; > //cout<<i<<" "<<j<<endl; > pp=dfs(i,j,0); > if(pp==50) > { > ar=i; > ac=j; > > break; > } > k+=2; > } > > > if(ar>=0&&ac>=0) > { > printf("%s found: (%d,%d)",ss,ar+1,ac+1); > for(i=0;i<strlen(ss)-1;i++) > { > switch(arr[i]) > { > case 0: > printf(", UL"); > break; > case 1: > printf(", U"); > break; > case 2: > printf(", UR"); > break; > case 3: > printf(", L"); > break; > case 4: > printf(", R"); > break; > case 5: > printf(", DL"); > break; > case 6: > printf(", D"); > break; > case 7: > printf(", DR"); > break; > case 8: > printf(", *"); > break; > > > } > } > printf("\n"); > } > else > printf("%s not found\n",ss); > } > } > > return 0; > } > > > waiting for ur reply.. > thanks :) > > regards > Anup Singh > > -- > You received this message because you are subscribed to the Google Groups > "Google Code Jam" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To post to this group, send email to [email protected]. > To view this discussion on the web visit > https://groups.google.com/d/msg/google-code/-/8YbMY3tD5-gJ. > For more options, visit https://groups.google.com/groups/opt_out. > > -- You received this message because you are subscribed to the Google Groups "Google Code Jam" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. For more options, visit https://groups.google.com/groups/opt_out.
