chase 300+200那种收到的信coupon,换10个包子# Money - 海外理财
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走迷宫的 时间复杂度是多少？谢谢 如这个解法
#include
// Maze size
#define N 4
bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N]);
/* A utility function to print solution matrix sol[N][N] */
void printSolution(int sol[N][N])
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
printf(" %d ", sol[i][j]);
printf("n");
}
}
/* A utility function to check if x,y is valid index for N*N maze */
bool isSafe(int maze[N][N], int x, int y)
{
// if (x,y outside maze) return false
if(x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)
return true;
return false;
}
/* This function solves the Maze problem using Backtracking. It mainly uses
solveMazeUtil() to solve the problem. It returns false if no path is
possible,
otherwise return true and prints the path in the form of 1s. Please note
that
there may be more than one solutions, this function prints one of the
feasible
solutions.*/
bool solveMaze(int maze[N][N])
{
int sol[N][N] = { {0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0}
};
if(solveMazeUtil(maze, 0, 0, sol) == false)
{
printf("Solution doesn't exist");
return false;
}
printSolution(sol);
return true;
}
/* A recursive utility function to solve Maze problem */
bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N])
{
// if (x,y is goal) return true
if(x == N-1 && y == N-1)
{
sol[x][y] = 1;
return true;
}
// Check if maze[x][y] is valid
if(isSafe(maze, x, y) == true)
{
// mark x,y as part of solution path
sol[x][y] = 1;
/* Move forward in x direction */
if (solveMazeUtil(maze, x+1, y, sol) == true)
return true;
/* If moving in x direction doesn't give solution then
Move down in y direction */
if (solveMazeUtil(maze, x, y+1, sol) == true)
return true;
/* If none of the above movements work then BACKTRACK:
unmark x,y as part of solution path */
sol[x][y] = 0;
return false;
}
return false;
}
// driver program to test above function
int main()
{
int maze[N][N] = { {1, 0, 0, 0},
{1, 1, 0, 1},
{0, 1, 0, 0},
{1, 1, 1, 1}
};
solveMaze(maze);
getchar();
return 0;
}
#include
// Maze size
#define N 4
bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N]);
/* A utility function to print solution matrix sol[N][N] */
void printSolution(int sol[N][N])
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
printf(" %d ", sol[i][j]);
printf("n");
}
}
/* A utility function to check if x,y is valid index for N*N maze */
bool isSafe(int maze[N][N], int x, int y)
{
// if (x,y outside maze) return false
if(x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)
return true;
return false;
}
/* This function solves the Maze problem using Backtracking. It mainly uses
solveMazeUtil() to solve the problem. It returns false if no path is
possible,
otherwise return true and prints the path in the form of 1s. Please note
that
there may be more than one solutions, this function prints one of the
feasible
solutions.*/
bool solveMaze(int maze[N][N])
{
int sol[N][N] = { {0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0}
};
if(solveMazeUtil(maze, 0, 0, sol) == false)
{
printf("Solution doesn't exist");
return false;
}
printSolution(sol);
return true;
}
/* A recursive utility function to solve Maze problem */
bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N])
{
// if (x,y is goal) return true
if(x == N-1 && y == N-1)
{
sol[x][y] = 1;
return true;
}
// Check if maze[x][y] is valid
if(isSafe(maze, x, y) == true)
{
// mark x,y as part of solution path
sol[x][y] = 1;
/* Move forward in x direction */
if (solveMazeUtil(maze, x+1, y, sol) == true)
return true;
/* If moving in x direction doesn't give solution then
Move down in y direction */
if (solveMazeUtil(maze, x, y+1, sol) == true)
return true;
/* If none of the above movements work then BACKTRACK:
unmark x,y as part of solution path */
sol[x][y] = 0;
return false;
}
return false;
}
// driver program to test above function
int main()
{
int maze[N][N] = { {1, 0, 0, 0},
{1, 1, 0, 1},
{0, 1, 0, 0},
{1, 1, 1, 1}
};
solveMaze(maze);
getchar();
return 0;
}
