C演算法 BFS（廣度優先搜尋法）

leetcode490題，迷宮，求到某個點是否存在路徑

leetcode505題，迷宮ii，求到某個點的最小路徑count

leetcode994題，腐爛的橘子，求各自中沒有新鮮橘子經歷的最小分鐘數。

演算法核心就是

1、寫好入隊函式的判斷

2、寫好外面函式入隊的場景

3、二維陣列需要構造結構體包含x，y座標

迷宮我用maze[x][y]==2標識這個區域是否被訪問過

迷宮ii我用maze[x][y]來標識到達當前位置所花的時間

迷宮

#define maxlen 10000

typedef
struct
queue;
queue g_queue[maxlen]
;int
enqueue
(int x,
int y,
int tail,
int*
*maze)
else
return tail +1;
}bool haspath
(int
** maze,
int mazesize,
int* mazecolsize,
int* start,
int startsize,
int* destination,
int destinationsize)
if(start[0]
== destination[0]
&& start[1]
== destination[1]
) row = mazesize;
col =
*mazecolsize;
dx = destination[0]
; dy = destination[1]
; head =0;
tail =0;
g_queue[tail]
.x = start[0]
; g_queue[tail]
.y = start[1]
; maze[g_queue[tail]
.x][g_queue[tail]
.y]=2;
tail++
;int tmpx, tmpy, count;
while
(head < tail)
while
(tmpx -
1>=
0&& maze[tmpx -1]
[tmpy]!=1
) tail =
enqueue
(tmpx, tmpy, tail,maze)
; tmpx = g_queue[head]
.x;while
(tmpx +
1< row && maze[tmpx +1]
[tmpy]!=1
) tail =
enqueue
(tmpx, tmpy, tail,maze)
; tmpx = g_queue[head]
.x;while
(tmpy -
1>=
0&& maze[tmpx]
[tmpy -1]
!=1) tail =
enqueue
(tmpx, tmpy, tail,maze)
; tmpy = g_queue[head]
.y;while
(tmpy +
1< col && maze[tmpx]
[tmpy +1]
!=1) tail =
enqueue
(tmpx, tmpy,tail,maze)
; head++;}
return false;
}

#define maxlen 10000

typedef
struct
queue;
queue g_queue[maxlen]
;int
enqueue
(int x,
int y,
int count,
int tail,
int*
*maze)
else
//printf("x=%d y=%d count=%d premaze=%d \n", x, y, count, maze[x][y]);
return tail +1;
}int
shortestdistance
(int
** maze,
int mazesize,
int* mazecolsize,
int* start,
int startsize,
int* destination,
int destinationsize)
if(start[0]
== destination[0]
&& start[1]
== destination[1]
) row = mazesize;
col =
*mazecolsize;
dx = destination[0]
; dy = destination[1]
; head =0;
tail =0;
g_queue[tail]
.x = start[0]
; g_queue[tail]
.y = start[1]
; maze[g_queue[tail]
.x][g_queue[tail]
.y]=2;
tail++
;int tmpx, tmpy, count;
while
(head < tail)
tail =
enqueue
(tmpx, tmpy, count, tail, maze)
; tmpx = g_queue[head]
.x; count = maze[tmpx]
[tmpy]
;while
(tmpx +
1< row && maze[tmpx +1]
[tmpy]!=1
) tail =
enqueue
(tmpx, tmpy, count, tail, maze)
; tmpx = g_queue[head]
.x; count = maze[tmpx]
[tmpy]
;while
(tmpy -
1>=
0&& maze[tmpx]
[tmpy -1]
!=1) tail =
enqueue
(tmpx, tmpy, count, tail, maze)
; tmpy = g_queue[head]
.y; count = maze[tmpx]
[tmpy]
;while
(tmpy +
1< col && maze[tmpx]
[tmpy +1]
!=1) tail =
enqueue
(tmpx, tmpy, count, tail, maze)
; head++;}
if(maze[dx]
[dy]==0
)return maze[dx]
[dy]-2
;}

#define maxlen 102

typedef
struct
queue;
queue q[maxlen]
;int
enqueue
(int i,
int j,
int row,
int col,
int*
* grid,
int tail)
if(grid[i]
[j]!=1)
grid[i]
[j]=2;
q[tail]
.x = i;
q[tail]
.y = j;
return tail +1;
}int
orangesrotting
(int
** grid,
int gridsize,
int* gridcolsize)}}
int number, min;
min =0;
while
(head < tail)
if(head < tail)
}for
(i =
0; i < row; i++)}
}return min;
}

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