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poj 2251(广搜求两点之间的距离)

2024-07-20 10:33:41   220人阅读

题目链接：http://poj.org/problem?id=2251

Dungeon Master
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16682 Accepted: 6491

Description

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides. 

Is an escape possible? If yes, how long will it take? 

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). 
L is the number of levels making up the dungeon. 
R and C are the number of rows and columns making up the plan of each level. 
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a ‘#‘ and empty cells are represented by a ‘.‘. Your starting position is indicated by ‘S‘ and the exit by the letter ‘E‘. There‘s a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form 
Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape. 
If it is not possible to escape, print the line 
Trapped!

Sample Input

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

Source

Ulm Local 1997
题解：（1）这道题比较简单，但是比较繁琐，要注意的细节比较多，=。=

             （2）广搜用的队列，求两点之间的距离一定是最短的~

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<string>
#include<stack>
#include<cmath>
#include<cctype>
#include<iostream>
#include<set>
#include<algorithm>
#include<ctime>
#include<vector>
using namespace std;
#define judge(x,y,z) !vis[x][y][z]&&map[x][y][z]!='#'&&x>=1&&x<=l&&y>=1&&y<=m&&z>=1&&z<=n

//设置方向数组优化广搜
int xx[]={0,0,0,0,1,-1};
int yy[]={-1,1,0,0,0,0};
int zz[]={0,0,-1,1,0,0};

bool vis[35][35][35];
char map[35][35][35];//地图
int l,m,n;

//设置结构体
struct node
{
    int x,y,z;
    int step;
};

int  BFS(int x,int y,int z)
{
    memset(vis,0,sizeof(vis));
    queue<node>q;
    node u;
    u.x=x;
    u.y=y;
    u.z=z;
    u.step=0;

    q.push(u);
    while(!q.empty())
    {
        u=q.front();
        if(map[u.x][u.y][u.z]=='E')return u.step;
        q.pop();
        for(int i=0;i<6;i++)
        {
            node v;
            v.x=u.x+xx[i];
            v.y=u.y+yy[i];
            v.z=u.z+zz[i];

            if(judge(v.x,v.y,v.z))
            {
                vis[v.x][v.y][v.z]=true;
                v.step=u.step+1;
                q.push(v);
            }
        }
    }
    return 0;
}
int main()
{
    while(cin>>l>>m>>n)
    {
        memset(map,0,sizeof(map));
        if(!m&&!n&&!l)break;
        int sx,sy,sz;
        for(int i=1;i<=l;i++)
          for(int j=1;j<=m;j++)
           for(int k=1;k<=n;k++)
           {
             cin>>map[i][j][k]; //这里注意一下：表示第i层第j行第k列
             if(map[i][j][k]=='S')
             {
                sx=i;
                sy=j;
                sz=k;
             }
           }

            int step=BFS(sx,sy,sz);

        if(step)printf("Escaped in %d minute(s).\n",step);
        else printf("Trapped!\n");
    }
    return 0;
}


poj 2251(广搜求两点之间的距离)


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