Dungeon Master
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 50464 Accepted: 18983
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!
题意：
给你一个三维的矩阵，有开始点，有结束点，每次可以走东、西、南、北、上、下。
其中有些点为石头，不能通过，现在求从开始点走到结束点的最短时间。如果不能走到则输出"Trapped!"。
思路：
一开始用的DFS,超时了。改成BFS就好了，直接套模板即可。。。
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 32;
char maze[MAXN][MAXN][MAXN];
bool book[MAXN][MAXN][MAXN];
int x2, y2, z2;
int move[6][3] = { {0,0,1}, {0,1,0}, {0,-1,0}, {0,0,-1}, {1,0,0}, {-1,0,0}};
int Min, L, R, C;;
struct State
{
int x, y, z;
int Step;
};
bool check(State s)
{
if(s.x >= L || s.x < 0 || s.y >= R || s.y < 0 || s.z >= C || s.z < 0)
return 0;
if(book[s.x][s.y][s.z] == false && maze[s.x][s.y][s.z] != '#') {
return 1;
}
return 0;
}
void bfs(State s)
{
queue <State> q;
State now, next;
q.push(s);
book[s.x][s.y][s.z] = true;
while(!q.empty())
{
now = q.front();
if(now.x == x2 && now.y == y2 && now.z == z2){
Min = now.Step;
return ;
}
for(int i = 0; i < 6; i++) {
next.x = now.x + move[i][0];
next.y = now.y + move[i][1];
next.z = now.z + move[i][2];
if(check(next)) {
next.Step = now.Step + 1;
q.push(next);
book[next.x][next.y][next.z] = true;
}
}
q.pop();
}
return ;
}
int main()
{
while(scanf("%d%d%d", &L, &R, &C)!=EOF)
{
if(L == 0 && R == 0 && C == 0)
break;
memset(book, false, sizeof(book));
int x1, y1, z1;
for(int i = 0; i < L; i++)
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') {
x1 = i;y1 = j; z1 = k;
}
if(maze[i][j][k] == 'E') {
x2 = i; y2 = j; z2 = k;
}
}
}
State s;
s.x = x1; s.y = y1; s.z = z1; s.Step = 0;
Min = 9999999;
bfs(s);
if(Min == 9999999)
printf("Trapped!\n");
else
printf("Escaped in %d minute(s).\n", Min);
}
return 0;
}
