标签：geo replace can bfs position pos cap trap gen
描述
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?
输入
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.
输出
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.
Trapped!
样例输入
3 4 5
S....
.###.
.##..
###.#
#####
#####
##.##
##...
#####
#####
#.###
####E
1 3 3
S##
#E#
###
0 0 0
样例输出
Escaped in 11 minute(s).
Trapped!
#include<bits/stdc++.h> using namespace std; char G[35][35][35]; int h,r,c,minn; struct node{ int x,y,z,s; }a; void bfs(int i,int j,int k,int s) { queue<node>qu; qu.push(node{i,j,k,0}); while(!qu.empty()) { a=qu.front(); qu.pop(); if(a.x>=1&&a.y>=1&&a.z>=1&&a.x<=h&&a.y<=r&&a.z<=c&&G[a.x][a.y][a.z]!=‘#‘) { qu.push(node{a.x+1,a.y,a.z,a.s+1}); qu.push(node{a.x-1,a.y,a.z,a.s+1}); qu.push(node{a.x,a.y+1,a.z,a.s+1}); qu.push(node{a.x,a.y-1,a.z,a.s+1}); qu.push(node{a.x,a.y,a.z+1,a.s+1}); qu.push(node{a.x,a.y,a.z-1,a.s+1}); } if(G[a.x][a.y][a.z]==‘E‘) { minn=a.s; return; } G[a.x][a.y][a.z]=‘#‘; } return; } int main() { int i,j,k,flag; while(scanf("%d%d%d",&h,&r,&c)!=EOF,h||r||c) { minn=99999999; flag=0; for(i=1;i<=h;i++) for(j=1;j<=r;j++) scanf("%s",G[i][j]+1); for(i=1;i<=h;i++) for(j=1;j<=r;j++) for(k=1;k<=c;k++) if(G[i][j][k]==‘S‘) { bfs(i,j,k,0); flag=1; break; } if(minn==99999999) printf("Trapped!\n"); else printf("Escaped in %d minute(s).\n",minn); } return 0; }
标签：geo replace can bfs position pos cap trap gen
原文地址：https://www.cnblogs.com/kannyi/p/8998317.html
