Uva532(三维搜索)

题目链接https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=473

= =这题就是就是给你一个三维的图，然后给你一个起点，一个终点，中间有障碍= =问你是否能够从起点走到终点，如果能最少的步数是多少

最优解= =用bfs，因为是三维，所以用三个方向数组来来确定下一步走哪里

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue> 
#include<algorithm>
#define N 33
using namespace std;
int f[N][N][N];
int x[6]={1,-1,0,0,0,0},y[6]={0,0,1,-1,0,0},z[6]={0,0,0,0,-1,1};
struct node
{
    int x,y,z,step;
};
int sx,sy,sz,ex,ey,ez;
int n,m,k;
int flag;
int bfs()
{
    queue<node>q;
    node in,out;
    in.x=sx,in.y=sy,in.z=sz;
    f[sx][sy][sz]=1;
    in.step=0;
    q.push(in);    
    while(!q.empty())
    {
        in=q.front();
        q.pop(); 
        if(in.x==ex&&in.y==ey&&in.z==ez)
        {
        flag=1;
        return in.step;
        }
        for(int i=0;i<6;i++)
        {
            out.x=in.x+x[i];
            out.y=in.y+y[i];
            out.z=in.z+z[i];
            out.step=in.step; 
            if(out.x>=0&&out.x<n&&out.y>=0&&out.y<m&&out.z>=0&&out.z<k&&!f[out.x][out.y][out.z])
            { 
            f[out.x][out.y][out.z]=1; 
            out.step++;   
            q.push(out); 
            } 
        }
    }
}
int main()
{
    while(scanf("%d%d%d",&k,&n,&m)!=EOF&&n&&m&&k)
    {
        flag=0;
        char s[100];
        memset(f,0,sizeof(f));
        for(int i=0;i<k;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%s",s);
                for(int l=0;l<m;l++)
                {
                    if(s[l]=='S')
                    {
                    sx=j;
                    sy=l;
                    sz=i;
                    }
                    else if(s[l]=='E')
                    {
                    ex=j;
                    ey=l;
                    ez=i;
                    }
                    else if(s[l]=='#')
                    f[j][l][i]=1;
                }
            }
        }
        int ans=bfs();
        if(flag)
        printf("Escaped in %d minute(s).\n",ans);
        else
        printf("Trapped!\n");
    }
    return 0;
}