HDU-1010-Tempter of the bone

Question Link

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1010

The question is that a dog eats bones, and it enters a maze trap. Every time a floor passes through the maze, the floor will collapse in the next second, so you cannot stay on the floor. There is a door in the maze that can only be opened at a specific second to let the dog escape. Now the question tells you the size of the maze and the time when the door is opened. If you ask your dog if you can escape, you can output yes; otherwise, no.

 

Pruning used for search:

1. If the current time is the number of steps (STEP)> = T and no dpoint is found, cut it.

2. set the shortest distance from the current location (x, y) to the dpoint (dx, Dy) to S. It takes time (Steps) to reach the current location (x, y, if the time required by the question is T-step <s, it is cut off.

3. For the current position (x, y), if (t-step-S) is an odd number, it is cut (parity pruning ).

4. If the number of points that can be taken in the map (xnum) <required time t, then cut (path pruning)

 

# Include <stdio. h> # Include <stdlib. h> Int n, m, ex, ey, T; Int success; Char maze [10] [10]; Void DFS (int stx, int sty, int DT) { If (Stx <= 0 | STX> N | sty <= 0 | sty> m) Return; If (Stx = ex & sty = ey & dt = T) { Success = 1; Return; } Int temp = (t-Dt)-ABS (ex-STX)-ABS (ey-STY ); If (temp <0 | temp & 1) Return; If (maze [STX] [sty + 1]! = 'X ') { Maze [STX] [sty + 1] = 'X '; DFS (Stx, sty + 1, DT + 1 ); Maze [STX] [sty + 1] = '.'; } If (maze [STX] [sty-1]! = 'X ') { Maze [STX] [sty-1] = 'X '; DFS (Stx, Sty-1, DT + 1 ); Maze [STX] [sty-1] = '.'; } If (maze [STX + 1] [sty]! = 'X ') { Maze [STX + 1] [sty] = 'X '; DFS (Stx + 1, sty, DT + 1 ); Maze [STX + 1] [sty] = '.'; } If (maze [stx-1] [sty]! = 'X ') { Maze [stx-1] [sty] = 'X '; DFS (stx-1, sty, DT + 1 ); Maze [stx-1] [sty] = '.'; } }   Int main (void) { Int STX, sty, wall, I, J; While (scanf ("% d", & N, & M, & T) = 3 & (n + M + T )) { Getchar (); Wall = 0; For (I = 1; I <= N; I ++) { For (j = 1; j <= m; j ++) { Scanf ("% C", & maze [I] [J]); If (maze [I] [J] ='s ') { STX = I; Sty = J; } Else if (maze [I] [J] = 'D ') { Ex = I; Ey = J; } Else if (maze [I] [J] = 'X ') Wall ++; } Getchar (); } Success = 0; Maze [STX] [sty] = 'X '; If (N * m-wall <= T) Printf ("NO \ n "); Else { DFS (Stx, sty, 0 ); If (SUCCESS) Printf ("Yes \ n "); Else Printf ("NO \ n "); } } Return 0; }       What is parity pruning? Take the matrix as follows: 0 1 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 1 Take a step from a grid with 0 to a grid with 1. A step from a grid with 1 is bound to a grid with 0. That is: Moving from 0 to 1 must be an odd step, and moving from 0 to 0 must be an even step. Therefore, when the request is from 0 to 0 but requires an odd number of times or from 1 to 0 but requires an even number of times, you can directly determine that the request is not reachable!   For example, there is a map:   [C-sharp]View plaincopy S... .... .... .... ... D   The shortest distance from S to D is S = ABS (DX-SX) + ABS (dy-sy ). If a map contains obstacles that cannot pass through:   [C-sharp]View plaincopy S.. x XX. x ... X . Xxx ... D   At this time, the shortest distance s '= S + 4. to bypass the obstacle, no matter the offset points, the offset distance is the shortest distance s plus an even distance. Just like the matrix mentioned above, it requires you to go from 0 to 0. No matter how you go around, it will always be the shortest distance (even step) plus an even step. It requires you to go from 1 to 0, you can only add an even number to the shortest distance (odd step.     You can also search for parity pruning.

HDU-1010-Tempter of the bone