HDU 3638 go, Susu

Hdu_3638

This question is not much different from that of a common BFS in the maze. Therefore, you may wish to separate the monster from each other and process the view of the monster every second, then, this second will not go to the monster's field of view, but will not go to the * position.

When dealing with the view of a monster, you can first use a struct to store the monster's location, and automatically update the monster's location every second, create a table with four directions and nine relative positions in each direction, which is advantageous for the view of the monster.

# Include <stdio. h> # Include < String . H> # Include <Algorithm> # Include <Queue> # Define Maxn 60 Int Trans [] = {0 , 2 , 1 , 4 , 3  };  Int DX [] = { 0 ,- 1 , 1 , 0 , 0 }, Dy [] = { 0 , 0 ,0 ,- 1 , 1  };  Int Ex [] [ 9 ] = {{  0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 ,0  },{  0 ,- 1 ,- 1 ,- 1 ,- 2 ,- 2 ,- 2 ,- 2 ,- 2  },{  0 , 1 , 1 ,1 , 2 , 2 , 2 , 2 , 2  },{  0 ,- 1 , 0 , 1 ,- 2 ,- 1 , 0 , 1 , 2  },{  0 ,- 1 , 0 , 1 ,- 2 ,- 1 , 0 , 1 , 2  }};  Int Ey [] [ 9 ] = {{ 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0  },{  0 ,- 1 , 0 , 1 ,- 2 ,- 1 , 0 , 1 , 2  },{  0 ,- 1 , 0 , 1 ,- 2 ,- 1 , 0 , 1 , 2  },{ 0 ,- 1 ,- 1 ,- 1 ,- 2 ,- 2 ,- 2 ,- 2 ,- 2  },{  0 , 1 , 1 , 1 , 2 , 2 , 2 , 2 , 2  }};  Struct  Monster {  Int  X, Y, D;} mon [maxn];  Char  B [maxn];  Int  N, m, K, Sx, Sy, TX, Ty, G [maxn] [maxn], del [maxn] [maxn], vis [maxn] [maxn];  Struct  Point { Int  X, Y, T; point () {} Point (  Int _ X, Int _ Y, Int  _ T): X (_ x), y (_ y), T (_ t) {}}; inline  Int Inside ( Int X, Int  Y ){  Return X> = 1 & X <= N & Y> = 1 & Y <=M ;}  Void  Moveall (){  Int  I, X, Y;  For (I = 0 ; I <K; I ++ ) {X = Mon [I]. x + dx [Mon [I]. d], y = mon [I]. Y + Dy [Mon [I]. d];  If (Inside (x, y) & G [x] [Y]) Mon [I]. x = x, mon [I]. Y = Y;  Else Mon [I]. d = Trans [Mon [I]. d];} Void  Delblock (){  Int  I, j, X, Y;  For (I = 1 ; I <= N; I ++ )  For (J = 1 ; J <= m; j ++) del [I] [J] = 1 - G [I] [J];  For (I = 0 ; I <K; I ++ ) For (J = 0 ; J < 9 ; J ++ ) {X = Mon [I]. x + ex [Mon [I]. d] [J], y = mon [I]. Y + Ey [Mon [I]. d] [J];  If (Inside (x, y) del [x] [Y] = 1  ;}}  Void  Init (){  Int  I, J; scanf (  "  % D " , & N ,& M );  For (I = 1 ; I <= N; I ++ ) {Scanf (  "  % S  " , B + 1  );  For (J = 1 ; J <= m; j ++ ) {G [I] [J] = B [J]! = ' *  '  ;  If (B [J] = '  A  ' ) SX = I, Sy = J;  Else   If (B [J] = '  B  ' ) Tx = I, Ty = J ;}} scanf (  "  % D " ,& K );  For (I = 0 ; I <K; I ++) scanf ( "  % D  " , & Mon [I]. X, & mon [I]. Y ,& Mon [I]. d );}  Void  Solve (){  Int I, j, X, Y, cur, ANS =- 1  ; Cur =- 1 ; Delblock ();  If  (DEL [SX] [sy]) {printf (  "  The victory and defeat of the military are not coming back soon.  "  );  Return  ;} STD: queue <Point> Q; q. Push (point (sx, Sy,  0  ));  While (! Q. Empty () {point P =Q. Front (); q. Pop ();  If (P. T> 1000 ) Break  ;  If (P. x = TX & P. Y = Ty) {ans = P. t;  Break  ;}  If (P. T> Cur) {cur = P. t; For (I = 1 ; I <= N; I ++ )  For (J = 1 ; J <= m; j ++) vis [I] [J] = 0  ; Moveall (), delblock ();}  For (I = 0 ; I < 5 ; I ++ ) {X = P. x + dx [I], y = P. Y + Dy [I]; If (Inside (x, y )&&! Del [x] [Y] &! Vis [x] [Y]) vis [x] [Y] = 1 , Q. Push (point (X, Y, P. t + 1  ));}}  If (ANS =- 1 ) Printf ( "  The victory and defeat of the military are not coming back soon.  "  );  Else Printf ( " % D \ n  "  , ANS );}  Int  Main (){  Int  T, TT; scanf (  "  % D  " ,& T );  For (Tt = 1 ; TT <= T; TT ++ ) {Init (); printf (  " Case % d:  "  , TT); solve ();}  Return   0  ;}