题目:http://poj.org/problem?id=2251
题意：输入一个L层的R*C迷宫，从S点出发，判断能否到达E，如果能，输出最短时间。
解法：简单bfs，把四个方向向量换成六个即可。
AC：
#include <iostream> #include <stdio.h> #include <string.h> #include <queue> #include <algorithm> using namespace std; char map[35][35][35]; int vis[35][35][35]; int k,n,m,sx,sy,sz,ex,ey,ez; int to[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,step; }; int check(int x,int y,int z) { if(x<0 || y<0 || z<0 || x>=k || y>=n || z>=m) return 1; else if(map[x][y][z] == '#') return 1; else if(vis[x][y][z]) return 1; return 0; } int bfs() { int i; node a,next; queue<node> Q; a.x = sx,a.y = sy,a.z = sz; a.step = 0; vis[sx][sy][sz] = 1; Q.push(a); while(!Q.empty()) { a = Q.front(); Q.pop(); if(a.x == ex && a.y == ey && a.z == ez) return a.step; for(i = 0; i<6; i++) { next = a; next.x = a.x+to[i][0]; next.y = a.y+to[i][1]; next.z = a.z+to[i][2]; if(check(next.x,next.y,next.z)) continue; vis[next.x][next.y][next.z] = 1; next.step = a.step+1; Q.push(next); } } return 0; } int main() { int i,j,r; while(scanf("%d%d%d",&k,&n,&m),n+m+k) { for(i = 0; i<k; i++) { for(j = 0; j<n; j++) { scanf("%s",map[i][j]); for(r = 0; r<m; r++) { if(map[i][j][r] == 'S') { sx = i,sy = j,sz = r; } else if(map[i][j][r] == 'E') { ex = i,ey = j,ez = r; } } } } memset(vis,0,sizeof(vis)); int ans; ans = bfs(); if(ans) printf("Escaped in %d minute(s).\n",ans); else printf("Trapped!\n"); } return 0; }