POJ 2251 Dungeon Master【三维BFS模板】

标签：algorithm   搜索   lan   empty   访问   miss   []   void   sed   

Dungeon Master
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 45743 Accepted: 17256
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

【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 1e6;
const int maxm = 100;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int dir[6][3]={ {0,0,1},{0,0,-1},{-1,0,0},{1,0,0},{0,1,0},{0,-1,0} };
struct node
{
    int x;
    int y;
    int z;
    int step;
}st,ed;
char maze[50][50][50]; //迷宫
int vis[50][50][50];   //表示是否访问过
int l,r,c; //l表示层数 r表示行数 c表示列数
int sx,sy,sz;  //起点坐标
int ex,ey,ez;

int check(int z,int x,int y) //判断该点是否可以走
{
    if(x>=0&&x<r&&y>=0&&y<c&&z>=0&&z<l&&maze[z][x][y]!='#')
        return 1;
    else
        return 0;
}

void bfs()
{
    memset(vis,0,sizeof(vis));
    queue <node> q;
    //把起点放入队列，并且标记
    //node st,ed;
    st.x = sx;
    st.y = sy;
    st.z = sz;
    st.step = 0;
    vis[st.z][st.x][st.y]=1;
    q.push(st);
    //不断循环知道队列的长度为0
    while(!q.empty())
    {
        //从队列的最前端取除元素看是否有路可通，如果没有路则返回这个点
        st = q.front();
        q.pop();
        //如果取出来是终点，则证明逃出地牢 搜索结束
        if(maze[st.z][st.x][st.y]=='E')
        {
            printf("Escaped in %d minute(s).\n",st.step);
            return;
        }
        //六个方向遍历
        for(int i = 0 ; i < 6 ;i++)
        {
            ed.z = st.z + dir[i][0];
            ed.x = st.x + dir[i][1];
            ed.y = st.y + dir[i][2];
            ed.step = st.step+1;
            if(check(ed.z,ed.x,ed.y)==0) //如果此方向不通就继续判断其他方向
                continue;
            if(vis[ed.z][ed.x][ed.y]==0)
            {
                vis[ed.z][ed.x][ed.y] = 1;
                q.push(ed);
            }
        }
    }
    printf("Trapped!\n");
}

int main()
{
    while(scanf("%d%d%d",&l,&r,&c)&&l&&r&&c)
    {
        for(int i = 0; i < l; i++)//输入各个点的坐标
        {
            for(int j = 0; j < r; j++)
            {
                scanf("%s",maze[i][j]);
                for(int k = 0; k < c ; k++)
                {
                    if(maze[i][j][k]=='S')
                    {
                        sz = i;
                        sx = j;
                        sy = k;
                    }

                }
            }
        }

        bfs();
    }
}

POJ 2251 Dungeon Master【三维BFS模板】

标签：algorithm   搜索   lan   empty   访问   miss   []   void   sed   

原文地址：https://www.cnblogs.com/Roni-i/p/9192579.html