POJ-2251-Dungeon Master（BFS 优先队列）

B - Dungeon Master
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Status

Practice

POJ 2251
Appoint description:
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!

走过的路标记为#，时间少的先出队，自定义六个搜索方向，直至搜到出口或者队列为空

代码

#include<stdio.h>#include<algorithm>#include<iostream>#include<string.h>#include<queue>using namespace std;const int maxn=33;char map[maxn][maxn][maxn];int dis[6][3]= {{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};//搜索方向int star_x;int star_y;int star_z;//起始点坐标int end_x;int end_y;int end_z;int L,R,C;struct node{    int x;    int y;    int z;    int time;    friend bool operator <(node n1,node n2)    {        return n1.time>n2.time;    }};bool Judge(node end){    if(end.x>0&&end.x<=L&&end.y>0&&end.y<=R&&end.z>0&&end.z<=C&&map[end.x][end.y][end.z]!='#')//搜索坐标未越界,且可被搜索        return true;    return false;}void BFS(){    node star;    star.time=0;    star.x=star_x;    star.y=star_y;    star.z=star_z;    map[star.x][star.y][star.z]='#';    priority_queue<node>q;    q.push(star);    while(!q.empty())    {        star=q.top();        q.pop();        if(star.x==end_x&&star.y==end_y&&star.z==end_z)        {            printf("Escaped in %d minute(s).\n",star.time);            return;        }        for(int i=0; i<6; i++) //六个搜索方向        {            node end;            end.x=star.x+dis[i][0];            end.y=star.y+dis[i][1];            end.z=star.z+dis[i][2];            if(Judge(end))            {//                printf("*****\n");//                printf("%d %d %d\n",end.x,end.y,end.z);//打印搜索路径                end.time=star.time+1;                map[end.x][end.y][end.z]='#';                q.push(end);            }        }    }    printf("Trapped!\n");}int main(){    while(~scanf("%d%d%d",&L,&R,&C)&&L&&R&&C)    {        for(int i=1; i<=L; i++)        {            for(int j=1; j<=R; j++)            {                for(int k=1; k<=C; k++)                {                    cin>>map[i][j][k];                    if(map[i][j][k]=='S')                    {                        star_x=i;                        star_y=j;                        star_z=k;                    }                    if(map[i][j][k]=='E')                    {                        end_x=i;                        end_y=j;                        end_z=k;//                        printf("%d %d %d\n",i,j,k);                    }                }//                getchar();            }        }        BFS();    }    return 0;}