poj 2251 Dungeon Master

题目链接：Dungeon Master

题目大意：给你一个三维的空间，然后给你起点和终点，中间有墙壁不能访问，问是否能从起点访问到终点，如果能，输出最短路径

题目思路：水题，写一个简单的bfs就可以了

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#include <iostream>#include <queue>using namespace std;int k,n,m,sx,sy,sz,ex,ey,ez;char mp[35][35][35];int dir[6][3] = {{1,0,0},{-1,0,0},{0,-1,0},{0,1,0},{0,0,1},{0,0,-1}};bool vis[35][35][35];struct node{    int x,y,z,step;};bool check(int x,int y,int z){    if(x < 0||y < 0||z < 0||x >= k||y >= n||z >= m||vis[x][y][z]||mp[x][y][z] == '#') return true;    return false;}int bfs(){    node a,next;    queue<node>Q;    a.x = sx,a.y = sy,a.z = sz,a.step = 0;    vis[sx][sy][sz] = true;    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(int i = 0;i < 6;i++){            next = a;            next.x = a.x+dir[i][0];            next.y = a.y+dir[i][1];            next.z = a.z+dir[i][2];            if(check(next.x,next.y,next.z))                continue;            vis[next.x][next.y][next.z] = true;            next.step = a.step + 1;            Q.push(next);        }    }    return 0;}int main(){    while(~scanf("%d%d%d",&k,&n,&m),n+m+k){        for(int i = 0;i < k;i++){            for(int j = 0;j < n;j++){                scanf("%s",mp[i][j]);                for(int r = 0;r < m;r++){                    if(mp[i][j][r] == 'S')                        sx = i,sy = j,sz = r;                    if(mp[i][j][r] == 'E')                        ex = i,ey = j,ez = r;                }            }        }        memset(vis,false,sizeof(vis));        int ans = bfs();        if(ans != 0) printf("Escaped in %d minute(s).\n",ans);        else printf("Trapped!\n");    }    return 0;}