POJ2251 Dungeon Master 三维BFS

题目大意：在一个三维迷宫中，问你从起点到终点的最小步数。

分析：和HDU1253差不多，简单的三维BFS。

实现代码：

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int dir[6][3]={{1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1}};
int map[55][55][55];
bool s[55][55][55];
int a,b,c;
int cx,cy,cz;
int sx,sy,sz,ex,ey,ez;
typedef struct node
{
    int x,y,z,tmp;
}point;
point que[51*51*51];
bool go(int x,int y,int z)
{
    if(x<=0||x>a||y<=0||y>b||z<=0||z>c)
      return false;
    else return true;
}
void bfs(int sx,int sy,int sz)
{
    int x,y,z;
    int cmp1=1,cmp2=1;
    que[cmp1].x=sx;
    que[cmp1].y=sy;
    que[cmp1].z=sz;
    que[cmp1].tmp=0;
    s[sx][sy][sz]=true;
    while(cmp1<=cmp2&&!s[ex][ey][ez])
    {
        x=que[cmp1].x;
        y=que[cmp1].y;
        z=que[cmp1].z;
        for(int i=0;i<6;i++)
          if(map[x+dir[i][0]][y+dir[i][1]][z+dir[i][2]])
          {
              cx=x+dir[i][0];
              cy=y+dir[i][1];
              cz=z+dir[i][2];
              if(go(cx,cy,cz)&&!s[cx][cy][cz])
              {
                  cmp2++;
                  que[cmp2].x=cx;
                  que[cmp2].y=cy;
                  que[cmp2].z=cz;
                  que[cmp2].tmp=que[cmp1].tmp+1;
                  s[cx][cy][cz]=true;
              }
          }
        cmp1++;
    }
    if(cmp1>cmp2)  puts("Trapped!");
    else printf("Escaped in %d minute(s).\n",que[cmp2].tmp);
}
int main()
{
    int cas,i,j,k;
    char tmp;
    while(scanf("%d%d%d",&a,&b,&c))
    {
        if(a==0&&b==0&&c==0) break;
        memset(map,1,sizeof(map));
        for(i=1;i<=a;i++)
          for(j=1;j<=b;j++)
            for(k=1;k<=c;k++)
            {
                cin>>tmp;
                if(tmp=='#') map[i][j][k]=0;
                if(tmp=='S'){sx=i;sy=j;sz=k;}
                if(tmp=='E'){ex=i;ey=j;ez=k;}
            }
        memset(s,false,sizeof(s));
        bfs(sx,sy,sz);
    }
    return 0;
}