poj 3984 迷宫问题

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迷宫问题

Description

定义一个二维数组： 

int maze[5][5] = {0, 1, 0, 0, 0,0, 1, 0, 1, 0,0, 0, 0, 0, 0,0, 1, 1, 1, 0,0, 0, 0, 1, 0,};

它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短路线。

Input

一个5 × 5的二维数组，表示一个迷宫。数据保证有唯一解。

Output

左上角到右下角的最短路径，格式如样例所示。

Sample Input

0 1 0 0 00 1 0 1 00 0 0 0 00 1 1 1 00 0 0 1 0

Sample Output

(0, 0)(1, 0)(2, 0)(2, 1)(2, 2)(2, 3)(2, 4)(3, 4)(4, 4)

水过。。。。

#include<stdio.h>#include<stdlib.h>#include<string.h>#define  N 5#define INF 0x7fffffftypedef struct node{ int x, y; }Node;Node cur, now, fa[N][ N], Queue[N << 5];int maze[N][N], Drict[N][N],last_dir[N][N],dir[N * N];int dx[] = { -1, 1, 0, 0 }, dy[] = { 0, 0, -1, 1 };void read(){int i, j;for (i = 0; i < N; i++){for (j = 0; j < N; j++)scanf("%d", &maze[i][j]);}}void bfs(int x, int y){int i, j, nx, ny, rear = 0, front = 0;for (i = 0; i < N; i++){for (j = 0; j < N; j++)Drict[i][j] = INF;}Drict[x][x] = 0;cur.x = x, cur.y = x;Queue[rear++] = cur;while (front < rear){now = Queue[front++];if (y == now.x && y == now.y)break;for (i = 0; i < 4; i++){nx = now.x + dx[i], ny = now.y + dy[i];if (nx >= 0 && nx < N && ny >= 0 && ny < N && INF == Drict[nx][ny] && 1!= maze[nx][ny]){cur = (Node){ nx, ny };Drict[nx][ny] = Drict[now.x][now.y] + 1;fa[nx][ny] = now;Queue[rear++] = cur;last_dir[nx][ny] = i;}}}}void print_path(int x, int y){int i, c = 0;while (1){int fx = fa[x][y].x, fy = fa[x][y].y;if (fx == x && fy == y)break;dir[c++] = last_dir[x][y];x = fx, y = fy;}x = y = 0;printf("(0, 0)\n");while (c--){for (i = 0; i<4; i++){if (i == dir[c]){printf("(%d, %d)\n", x += dx[i], y += dy[i]);break;}}}}void solve(){memset(Queue, 0, sizeof(Queue));bfs(0, 4);print_path(4, 4);}int main(){read();solve();return 0;}