#include <stdio.h> #include <stdlib.h> typedef struct point { int x; int y; int distance; } Point; struct QueueRecord { int capacity ; /* max size of queue */ int front; /* the front pointer */ int rear; /* the rear pointer */ Point *base; /* array for queue elements */ } ; typedef struct QueueRecord *Queue; Queue CreateQueue (int MaxElements) { Queue Q = (Queue)malloc(sizeof(struct QueueRecord)); if (Q == NULL) printf ("Out of space!!!"); Q->base = (Point*)malloc(sizeof(Point) * MaxElements); if (Q->base == NULL) printf("Out of space!!!"); Q->capacity = MaxElements; Q->front = Q->rear = 0; return Q; } int IsEmpty( Queue Q ) { if (Q->front == Q->rear) return 1; return 0; } void Enqueue(Point e, Queue Q) { if ((Q->rear + 1) % Q->capacity == Q->front) { printf("Queue is full!!"); return; } Q->base[Q->rear] = e; Q->rear = (Q->rear + 1) % Q->capacity; } Point Dequeue (Queue Q) { Point e; if (IsEmpty(Q)) { printf("Queue is empty!!"); exit(0); } e = Q->base[Q->front]; Q->front = (Q->front + 1) % Q->capacity; return e; } Point Front (Queue Q) { Point e; if (IsEmpty(Q)) { printf("Queue is empty!!"); exit(0); } e = Q->base[Q->front]; return e; } int main() { Queue Q = CreateQueue(25); Point startpoint = {1, 1, 0}; Point endpoint = {5, 5, 0}; int data[7][7] = {0}; int i, j; for (i = 1; i <= 5; i++) for (j = 1; j <= 5; j++) scanf("%d", &data[i][j]); return 0; } 请补全代码给出一个5×5的矩阵构成的迷宫的地图,其中0为障碍, 1为可通行的地方。迷宫的入口为左上角(1, 1),出口为右下角(5, 5),在迷宫中,只能从一个位置走到它的上、下、左、右四个方向之一。对于输入的迷宫结构,请找到从迷宫入口到出口的最少步数并输出(若不存在从入口到出口的路径,则输出: oops!) 输入说明:整型元素0或者1构成的5×5矩阵 输出说明:代表通过迷宫最少步数的整数或者表示不存在路径的字符串"oops!
时间: 2023-12-03 13:42:22 浏览: 71
#include<stdio.h>
代码已经给出了基本的队列结构和一些函数,我们需要在main函数中补充完整代码来实现迷宫的广度优先搜索。
首先,我们需要将输入的矩阵转换成一个点的集合,其中每个点代表一个可通行的位置(即值为1),并在点的结构体中记录该点到起点的距离(distance)。具体实现如下:
```
// 将输入的矩阵转换成一个点的集合
Point points[25];
int pointIndex = 0;
for (i = 1; i <= 5; i++) {
for (j = 1; j <= 5; j++) {
if (data[i][j] == 1) {
Point p = {i, j, -1}; // 初始化距离为-1
points[pointIndex++] = p;
}
}
}
```
接下来,我们按照广度优先搜索的方法来查找从起点到终点的最短路径。具体实现如下:
```
// 将起点入队,并将其到起点的距离设置为0
startpoint.distance = 0;
Enqueue(startpoint, Q);
while (!IsEmpty(Q)) {
Point current = Dequeue(Q);
// 如果当前点是终点,则输出最短距离并退出循环
if (current.x == endpoint.x && current.y == endpoint.y) {
printf("%d", current.distance);
return 0;
}
// 否则,遍历当前点的四个方向
// 如果某个方向的点未被访问过,则将其入队,并更新到起点的距离
if (data[current.x-1][current.y] == 1 && points[(current.x-2)*5+current.y-1].distance == -1) {
Point next = {current.x-1, current.y, current.distance+1};
points[(current.x-2)*5+current.y-1].distance = current.distance+1;
Enqueue(next, Q);
}
if (data[current.x+1][current.y] == 1 && points[current.x*5+current.y-1].distance == -1) {
Point next = {current.x+1, current.y, current.distance+1};
points[current.x*5+current.y-1].distance = current.distance+1;
Enqueue(next, Q);
}
if (data[current.x][current.y-1] == 1 && points[(current.x-1)*5+current.y-2].distance == -1) {
Point next = {current.x, current.y-1, current.distance+1};
points[(current.x-1)*5+current.y-2].distance = current.distance+1;
Enqueue(next, Q);
}
if (data[current.x][current.y+1] == 1 && points[(current.x-1)*5+current.y].distance == -1) {
Point next = {current.x, current.y+1, current.distance+1};
points[(current.x-1)*5+current.y].distance = current.distance+1;
Enqueue(next, Q);
}
}
```
最后,如果循环结束后还没有找到终点,则说明不存在从起点到终点的路径,输出"oops!"即可。
完整代码如下:
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