UVA 532

题目大意：3D迷宫，’#‘为障碍物，’.‘为可行，’S‘为起点，’E‘为终点，求最短路径。

解题思路：bfs与迷宫问题类似，只是多一维，起点，终点坐标自己求一下。标记数组依然是步数数组。只要不混了，没什么难点。

ac代码：

#include <iostream>#include <queue>#include <cstring>using namespace std;struct node{int x;int y;int z;};queue <node> qu;int dx[6] = {0, 0, 0, 0, -1, 1};int dy[6] = {0, 0, -1, 1, 0, 0};int dz[6] = {-1, 1, 0, 0, 0, 0};int vis[100][100][100], n, m, p, ex[6], temp;char a[100][100][100];void search(){memset(vis, -1, sizeof(vis));for (int i=0; i<n; i++)for (int j=0; j<m; j++)for (int k=0; k<p; k++)if (a[i][j][k] == 'S')ex[0] = i, ex[1] = j, ex[2] = k;else if (a[i][j][k] == 'E')ex[3] = i, ex[4] = j, ex[5] = k;else if (a[i][j][k] == '#')vis[i][j][k] = 0;}int bfs(){while (!qu.empty())qu.pop();node no;no.x = ex[0], no.y = ex[1], no.z = ex[2];qu.push(no);vis[no.x][no.y][no.z] = 0;while (!qu.empty()){node temp = qu.front();qu.pop();int X = temp.x, Y = temp.y, Z = temp.z;if (X == ex[3] && Y == ex[4] && Z == ex[5])return vis[X][Y][Z];for (int i=0; i<6; i++){int a = X + dx[i], b = Y + dy[i];int c = Z + dz[i];if (a >=0 && a < n && b >= 0 && b < m && c >= 0 && c < p && vis[a][b][c] == -1){no.x = a, no.y = b, no.z = c;qu.push(no);vis[a][b][c] = vis[X][Y][Z] + 1; }}}return 0; }int main(){while (scanf("%d%d%d", &n, &m, &p)!=EOF){if (!n && !m && !p)break;for (int i=0; i<n; i++){for (int j=0; j<m; j++)scanf("%s", a[i][j]);getchar();getchar();}search();temp = bfs();if (temp)printf("Escaped in %d minute(s).\n", temp);elseprintf("Trapped!\n");}return 0;}