HDU 3681 Prison Break

Hdu_3681

Since y and G are less than 15, the shortest circuit between F, y, and g can be processed in advance and then converted into the TSP problem.

# Include <stdio. h> # Include < String . H> # Include <Algorithm> # Include <Queue> # Define Maxd 20 Int  N, m, DIS [maxd] [maxd], G [maxd] [maxd], Mark, Y, G, Sx, Sy, vis [maxd] [maxd];  Int DX [] = {-1 , 1 , 0 , 0 }, Dy [] = { 0 , 0 ,- 1 , 1  };  Char  B [maxd] [maxd];  Int F [ 1 < 15 | 10 ] [Maxd];  Void  Init (){  Int  I, J; y = 1  ;  For (I = 1 ; I <= N; I ++ ) {Scanf (  "  % S  " , B [I] + 1  );  For (J = 1 ; J <= m; j ++ ){  If (B [I] [J] = '  F  ' ) G [I] [J] = 0 , SX = I, Sy = J;  Else   If (B [I] [J] = '  S  ' ) G [I] [J] =- 1 ;  Else   If (B [I] [J] = '  Y  ' ) G [I] [J] = y ++ ;  Else   If (B [I] [J] = '  D  ' ) G [I] [J] =- 2  ;} Mark = ( 1 <Y )-2 , G = Y;  For (I = 1 ; I <= N; I ++ )  For (J = 1 ; J <= m; j ++ )  If (B [I] [J] = '  G  ' ) G [I] [J] = g ++ ;} Inline  Int Inside (Int X, Int  Y ){  Return X> = 1 & X <= N & Y> = 1 & Y <= M ;}  Void DFS ( Int X, Int  Y ){  Int  I, NX, NY; vis [x] [Y] = 1 ;  For (I = 0 ; I < 4 ; I ++ ) {NX = X + dx [I], NY = Y + Dy [I];  If (Inside (NX, NY )&&! Vis [NX] [NY] & G [NX] [NY]! =- 2  ) DFS (NX, NY );}}  Int  Check (){  Int  I, J; memset (VIS, 0 , Sizeof  (VIS); DFS (sx, Sy );  For (I = 1 ; I <= N; I ++ )  For (J = 1 ; J <= m; j ++ )  If (B [I] [J] = '  Y  ' &&! Vis [I] [J]) Return  0  ;  Return   1  ;}  Void BFS ( Int SX, Int  Sy ){  Int I, j, X, Y, NX, NY, id = G [SX] [sy]; memset (VIS, - 1 , Sizeof  (VIS); STD: queue < Int > Q; vis [SX] [sy] = 0  ; Q. Push (SX * (M + 1 ) + Sy );  While (! Q. Empty () {x = Q. Front ()/(m + 1 ), Y = Q. Front () % (m + 1  ), Q. Pop ();  If (G [x] [Y]> = 0 ) Dis [ID] [G [x] [Y] =Vis [x] [Y];  For (I = 0 ; I < 4 ; I ++ ) {NX = X + dx [I], NY = Y + Dy [I];  If (Inside (NX, NY) & Vis [NX] [NY] =- 1 & G [NX] [NY]! =- 2  ) Vis [NX] [NY] = Vis [x] [Y] + 1 , Q. Push (nx * (m + 1 ) +NY );}}}  Int DP ( Int  Limit ){  Int  I, J, K; memset (F, - 1 , Sizeof  (F); F [  1 ] [ 0 ] = Limit;  For (I = 1 ; I <( 1 <G); I ++ )  For (J = 0 ; J <G; j ++ ){  If (F [I] [J] =- 1 ) Continue  ;  If (I & mark) = mark) Return   1  ;  For (K = 0 ; K <G; k ++ )  If (I & 1 <K) = 0 & F [I] [J]> = Dis [J] [k]) {  If (K> = y) f [I | 1 <K] [k] = Limit;  Else F [I | 1 <K] [k] = STD: max (F [I | 1 <K] [K], F [I] [J]-Dis [J] [k]) ;}}  Return   0  ;}  Void  Solve (){  Int  I, J;  If (Y = 1  ) {Printf (  "  0 \ n  "  );  Return ;}  If (! Check () {printf (  "  -1 \ n  "  );  Return  ;} Memset (DIS,  0x3f , Sizeof  (DIS ));  For (I = 1 ; I <= N; I ++ ) For (J = 1 ; J <= m; j ++ )  If (G [I] [J]> = 0  ) BFS (I, j );  Int Min = 0 , Max = 1000  , Mid;  For  (;) {Mid = Min + MAX + 1 > 1 ;  If (Mid = max) Break  ;  If (Dp (MID) max = Mid;  Else Min = Mid;} printf (  "  % D \ n  "  , Mid );}  Int  Main (){  While (Scanf ("  % D  " , & N, & M), N | M) {Init (); solve ();}  Return   0  ;}