問題：
利用隊列的假出隊，實現(xiàn)迷官問題尋解。
要求：
0 墻 1 路， 尋找最短路， 如果有多條，輸出其中一條路。

main.c

#include <stdio.h> #include <stdlib.h>#define M 5 int map[M+2][M+2]; int front = 0, rear = 0; typedef struct Point_def{int x;int y;int pre; }Point; Point q[(M+2)*(M+2)]; typedef struct direction_def{int x;int y; }Direction; Direction direct[4] = {{-1,0}, {0,1}, {1,0}, {0, -1} }; int isEmpty(){return front==rear?1:0; } void initMap(){int i, j;for (i=0; i<M+2; i++)for (j=0; j<M+2; j++){if (i == 0 || i == M+1 || j == 0 || j == M+1){map[i][j] = 0; continue;}scanf("%d", &map[i][j]);} } void enQueue(Point p){q[rear++] = p; } Point deQueue(){if (isEmpty()) exit(1);return q[front++]; }void findPath(int start_x, int start_y, int end_x, int end_y){int flag = 0;int d;int i, j;Point tmp_p1={start_x, start_y, -1};Point tmp_p2;enQueue(tmp_p1);while (!flag){tmp_p1 = deQueue();map[tmp_p1.x][tmp_p1.y] = -1;for (d=0; d<4; d++){i = tmp_p1.x + direct[d].x;j = tmp_p1.y + direct[d].y;if (map[i][j] == 1){tmp_p2.x = i; tmp_p2.y = j; tmp_p2.pre = front-1;enQueue(tmp_p2);}if (i == end_x && j == end_y){tmp_p2.x = i; tmp_p2.y = j; tmp_p2.pre = front-1;enQueue(tmp_p2);flag = 1; break;}}// end for }// end while }// end findPathvoid printPath(){int k = rear-1;while (k != -1){printf("(%d,%d)\n", q[k].x, q[k].y);k = q[k].pre;} }int main(){int flag = 0;int start_x = 1, start_y = 1;int end_x = 5, end_y = 5;initMap();findPath(start_x, start_y, end_x, end_y);printPath();return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/laohaozi/p/8266613.html

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