一、内容
题意：给定一个3D的图，你能走6个方向，问从起点到终点最少的步数。
二、思路
- bfs搜索一下。
三、代码
#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
using namespace std;
int l, r, c, sl, sx, sy;
char g[35][35][35], vis[35][35][35];
int dx[6] = {-1, 1, 0, 0, 0, 0};
int dy[6] = {0, 0, -1, 1, 0, 0};
int dl[6] = {0, 0, 0, 0, 1, -1};
struct node {
int l, x, y, step;
node (int l, int x, int y, int step): l(l), x(x), y(y), step(step){}
};
bool ok(int fl, int fx, int fy) {
if (fl < 0 || fx < 0 || fy < 0 || fl >= l || fx >= r || fy >= c || g[fl][fx][fy] == '#') return false;
return true;
}
int bfs() {
queue<node> q;
q.push(node(sl, sx, sy, 0));
while (!q.empty()) {
node t = q.front();
q.pop();
if (g[t.l][t.x][t.y] == 'E') return t.step;
if (vis[t.l][t.x][t.y]) continue;
vis[t.l][t.x][t.y] = 1;
//6个方向
for (int i = 0; i < 6; i++) {
int fl = t.l + dl[i];
int fx = t.x + dx[i];
int fy = t.y + dy[i];
if (ok(fl, fx, fy)) {
q.push(node(fl, fx, fy, t.step + 1));
}
}
}
return -1;
}
int main() {
while (scanf("%d%d%d", &l, &r, &c), l) {
memset(vis, 0, sizeof(vis));
for (int i = 0; i < l; i++) {
for (int j = 0; j < r; j++) {
scanf("%s", g[i][j]);
}
}
for (int i = 0; i < l; i++) {
for (int j = 0; j < r; j++) {
for (int k = 0; k < c; k++) {
if (g[i][j][k] == 'S') {
sl = i, sx = j, sy = k;
break;
}
}
}
}
int step = bfs();
if (step == -1) {
printf("Trapped!\n");
} else {
printf("Escaped in %d minute(s).\n", step);
}
}
return 0;
}
