Data structure routine-Maze problem (using stack structure), data structure routine

Data structure routine-Maze problem (using stack structure), data structure routine

This article focuses on the data structure basic series of network courses (3): The application of the 6th Lesson stack in the stack and queue 2-Maze problem.

Example: Find the path from entry to exit

Program Implementation:

# Include <stdio. h> # define MaxSize 100 # define M 8 # define N 8int mg [M + 2] [N + 2] ={{ 1, 1, 1, 1, 1, 1, 1 }, {,}, {,}, {, 0,, ,}, {,}, {,}; typedef struct {int I; // The row number of the current square int j; // The column number int di of the current square; // di is the next azimuth number that can take the adjacent azimuth} Box; typedef struct {Box data [MaxSize]; int top; // stack top pointer} StType; // defines the stack type int mgpath (int xi, int yi, int xe, int ye) // The solution path is: (xi, yi)-> (xe, ye) {int I, j, k, di, find; StType st; // defines the stack st. top =-1; // initialize the stack top pointer st. top ++; // The initial block is pushed to the stack st. data [st. top]. I = xi; st. data [st. top]. j = yi; st. data [st. top]. di =-1; mg [xi] [yi] =-1; while (st. top>-1) // loop when the stack is not empty {I = st. data [st. top]. i; j = st. data [st. top]. j; di = st. data [st. top]. di; // get the top block of the stack if (I = xe & j = ye) // find the exit, and the output path {printf ("the maze path is as follows: \ n "); for (k = 0; k <= st. top; k ++) {printf ("\ t (% d, % d)", st. data [k]. i, st. data [k]. j); if (k + 1) % 5 = 0) // change the printf ("\ n") line after each of the five squares output ");} printf ("\ n"); return (1); // find a path and return 1} find = 0; while (di <4 & find = 0) // find the next square {di ++; switch (di) {case 0: I = st. data [st. top]. i-1; j = st. data [st. top]. j; break; case 1: I = st. data [st. top]. i; j = st. data [st. top]. j + 1; break; case 2: I = st. data [st. top]. I + 1; j = st. data [st. top]. j; break; case 3: I = st. data [st. top]. i, j = st. data [st. top]. j-1; break;} if (mg [I] [j] = 0) find = 1; // find the next adjacent square} if (find = 1) // found the next accessible square {st. data [st. top]. di = di; // modify the di value of the element at the top of the original stack. top ++; // the next block can be taken into the stack st. data [st. top]. I = I; st. data [st. top]. j = j; st. data [st. top]. di =-1; mg [I] [j] =-1; // avoid repeated steps to this square} else // if no path is available, the stack is rolled back {mg [st. data [st. top]. i] [st. data [st. top]. j] = 0; // change the location to another path to go to the square st. top --; // return the block to stack} return (0); // It indicates that no path is available, and 0} int main () {mgpath (, M, n); return 0 ;}

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