1
3 3 4 20
0 1 1 1
0 0 1 1
0 1 1 1
1 1 1 1
1 0 0 1
0 1 1 1
0 0 0 0
0 1 1 0
0 1 1 0

11

这一题是按照6个方向走的：上，下，左，右，前，后。创建一个三维数组记录当前的位置状态。


#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define N 60
int map[N][N][N];
int vis[N][N][N];
int t,a,b,c,tt;
int dx[] = {0,0,-1,1,0,0};
int dy[] = {0,0,0,0,-1,1};
int dz[] = {-1,1,0,0,0,0};
struct point
{
	int x,y,z,step;
};
int bfs()
{
	if (a - 1 == 0 && b - 1 == 0 && c - 1 == 0)//如果出口位置与入口相同，直接返回0.
		return 0;
	point temp, head, tt;
	queue<point>q;
	while (!q.empty())
		q.pop();
	vis[0][0][0] = 1;
	temp.x = 0; temp.y = 0; temp.z = 0; temp.step = 0;
	q.push(temp);
	while (!q.empty())
	{
		head = q.front();
		q.pop();
		for (int i = 0; i < 6; i++)
		{
			tt.x = head.x + dx[i];
			tt.y = head.y + dy[i];
			tt.z = head.z + dz[i];
			if (tt.x < 0 || tt.y < 0 || tt.z < 0 || tt.x >= a || tt.y >= b || tt.z >= c || vis[tt.x][tt.y][tt.z]||map[tt.x][tt.y][tt.z])
				continue;
			tt.step = head.step + 1;
			vis[tt.x][tt.y][tt.z] = 1;
			if (tt.x == a - 1 && tt.y == b - 1 && tt.z == c - 1)
				return tt.step;
			q.push(tt);
		}
	}
	return (-1);//如果找不到出口就返回-1.
}
int main()
{	
	scanf("%d", &t);
	while (t--)
	{
		memset(vis, 0, sizeof(vis));
		scanf("%d%d%d%d", &a, &b, &c, &tt);
		for (int i = 0; i < a; i++)
		{
			for (int j = 0; j < b; j++)
			{
				for (int k = 0; k < c; k++)
				{
					scanf("%d", &map[i][j][k]);
				}
			}
		}
		int ans = bfs();
		if (ans<0 || ans>tt)//返回值小于零找不到出口，大于魔王的时间都不能逃脱.
			printf("-1\n");
		else
			printf("%d\n", ans);
	}
	return 0;
}