//啊啊啊 终于被我搞定
//这个题写了快一下午
//第一次用的DFS 还是感觉DFS思路清晰
//但…超时 问了老师 这种题一般都要用BFS 我昨天做的 走迷宫 那个题 应该是比较小 我用了DFS
//这种题还是我第二次用BFS做呢
//现在我知道两个用BFS的模板了
其实和我昨天发的走迷宫真的是差不多 就是多了两个方向
套路一样
附上我在某位大侠那里看到的一个样例
输入
3 4 5
S....
.###.
.##..
###.#
#####
#####
##.##
##...
#####
#####
#.###
####E
1 3 3
S##
#E#
###
4 4 5
#####
#####
##.##
##...
#####
###S#
#.#.#
#....
#####
#####
#.###
####E
#####
#...#
...##
#....
1 2 2
##
SE
3 3 3
###
#E#
#.#
#..
.##
.#.
...
...
..S
5 2 2
##
##
#.
..
.S
..
..
..
.E
..
2 1 1
S
E
1 1 2
SE
1 5 5
#####
S####
....#
####E
##...
2 3 3
#.E
...
...
...
...
..S
2 4 4
####
#S##
....
###.
####
.###
.###
.E..
2 5 5
#####
#S..#
..#..
.###.
.###.
#####
#####
#####
#####
..E..
0 0 0
————————————————
原文链接：https://blog.csdn.net/HelloWorld10086/article/details/38323189
输出
Escaped in 11 minute(s).
Trapped!
Escaped in 4 minute(s).
Escaped in 1 minute(s).
Trapped!
Escaped in 2 minute(s).
Escaped in 1 minute(s).
Escaped in 1 minute(s).
Trapped!
Escaped in 3 minute(s).
Escaped in 5 minute(s).
Escaped in 7 minute(s).
BFS
#include <iostream>
#include <string>
#include <queue>
#include <cstring>
using namespace std;
string s[31][31];
int vis[31][31][31];
int L, n, m;
//arr 定位数组
int arr[6][3] = {{0, 0, 1}, {0, 0, -1}, {0, -1, 0}, {0, 1, 0}, {-1, 0, 0}, {1, 0, 0}};
struct node
{
int x, y, z, t;
};
int bfs(int x, int y, int z)
{
//t1 作为 之前的
//t2 就是 之后的
node t1, t2;
vis[x][y][z] = 1;
queue<node> q;
q.push({x, y, z, 0});
while (!q.empty())
{
t1 = q.front();
q.pop();
//6个走法
for (int i = 0; i < 6; i++)
{
t2.x = t1.x + arr[i][0];
t2.y = t1.y + arr[i][1];
t2.z = t1.z + arr[i][2];
t2.t = t1.t + 1;
if ((t2.x >= 0 && t2.x < L) && (t2.y >= 0 && t2.y < n) && (t2.z >= 0 && t2.z < m) && ((s[t2.x][t2.y][t2.z] == 'E' || s[t2.x][t2.y][t2.z] == '.') && !vis[t2.x][t2.y][t2.z]))
{
//如果正好到 E
if (s[t2.x][t2.y][t2.z] == 'E')
{
return t2.t;
}
vis[t2.x][t2.y][t2.z] = 1;
q.push(t2);
}
}
}
return -1;
}
int main()
{
while (1)
{
int ans;
cin >> L >> n >> m;
memset(vis, 0, sizeof(vis));
if (L == 0 && n == 0 && m == 0)
break;
//输入
for (int i = 0; i < L; i++)
{
for (int j = 0; j < n; ++j)
{
cin >> s[i][j];
}
}
//找入口
for (int i = 0; i < L; i++)
{
for (int j = 0; j < n; ++j)
{
for (int p = 0; p < m; ++p)
{
if (s[i][j][p] == 'S')
{
ans = bfs(i, j, p);
break;
}
}
}
}
if (ans == -1)
cout << "Trapped!";
else
printf("Escaped in %d minute(s).", ans);
cout << endl;
}
return 0;
}
DFS 超时代码
#include <iostream>
#include <vector>
#include <string>
using namespace std;
string s[100][100];
int L, n, m;
int ans = 1 << 30;
int arr[6][3] = {{0, 0, 1}, {0, 0, -1}, {0, -1, 0}, {0, 1, 0}, {-1, 0, 0}, {1, 0, 0}};
// x y z 是坐标 t是步数
void dfs(int x, int y, int z, int t)
{
if (t >= ans)
return;
//遍历6种走法
for (int i = 0; i < 6; i++)
{
int a, b, c;
a = x + arr[i][0];
b = y + arr[i][1];
c = z + arr[i][2];
if ((a >= 0 && a < L) && (b >= 0 && b < n) && (c >= 0 && c < m) && (s[a][b][c] == 'E' || s[a][b][c] == '.'))
{
//找到也就结束了
if (s[a][b][c] == 'E')
{
ans = min(ans, t);
return;
}
//恢复现场
s[x][y][z] = '*';
dfs(a, b, c, t + 1);
s[x][y][z] = '.';
}
}
}
int main()
{
cin.sync_with_stdio(false);
while (1)
{
ans = 1 << 30;
cin >> L >> n >> m;
if (L == 0 && n == 0 && m == 0)
break;
for (int i = 0; i < L; i++)
{
for (int j = 0; j < n; ++j)
{
cin >> s[i][j];
}
}
for (int i = 0; i < L; i++)
{
for (int j = 0; j < n; ++j)
{
for (int p = 0; p < m; ++p)
{
if (s[i][j][p] == 'S')
{
dfs(i, j, p, 1);
break;
}
}
}
}
if (ans == (1 << 30))
cout << "Trapped!";
else
cout << "Escaped in " << ans << " minute(" << 's' << ").";
cout << endl;
}
return 0;
}
来源：CSDN
作者：陌陌623
链接：https://blog.csdn.net/weixin_45653525/article/details/104117766