POJ 2251 Dungeon Master
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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!
这个题简直了。。。
好吧，其实这个题大意是一个３D迷宫，然后长宽高分别是Ｌ，Ｒ，Ｃ，然后有一个起点Ｓ有一个终点Ｅ，我们的目标是从Ｓ出发到达Ｅ，计算路程距离，假设移动一次距离加１。
这个就是把迷宫推广到三维空间，用BFS还是可以很轻松做出来，只要把方向稍作修改即可。
代码如下：
/*************************************************************************
> File Name: test_again.c
> Author: zhanghaoran
> Mail: 467908670@qq.com
> Created Time: 2015年06月01日 星期一 21时18分15秒
************************************************************************/
#include <stdio.h>
#include <string.h>
struct q{
int x;
int y;
int z;
}q[30000];
int length[30000];
int dx[] = {1, -1, 0, 0, 0, 0};
int dy[] = {0, 0, 0, 0, -1, 1};
int dz[] = {0, 0,- 1, 1, 0, 0};
int flag[40][40][40];
char map[40][40][40];
int L, R, C, nx, ny, nz;
int bfs(){
int rear,front;
int xx, yy, zz;
int i;
memset(flag, 0, sizeof(flag));
memset(length, 0, sizeof(length));
q[0].x = nx;
q[0].y = ny;
q[0].z = nz;
front = rear = 0;
while(front <= rear){
for(i = 0;i < 6;i ++){
xx = q[front].x + dx[i];
yy = q[front].y + dy[i];
zz = q[front].z + dz[i];
if(!flag[xx][yy][zz] && (map[xx][yy][zz] == '.' || map[xx][yy][zz] == 'E') && xx >= 0 && xx < L && yy >= 0 && yy < R && zz >= 0 && zz < C)
{
flag[xx][yy][zz] = 1;
q[++rear].x = xx;
q[rear].y = yy;
q[rear].z = zz;
length[rear] = length[front] + 1;
if(map[xx][yy][zz]=='E')
return length[rear];
}
}
front++;
}
return 0;
}
int main(){
int i, j, k, len;
while(scanf("%d%d%d\n", &L, &R, &C)){
if(L == 0 && R == 0 && C == 0)
break;
for(i = 0; i < L; i ++, getchar())
for(j = 0;j < R; j ++, getchar())
for(k = 0; k < C; k ++){
scanf("%c", &map[i][j][k]);
if(map[i][j][k] == 'S') {
nx = i;
ny = j;
nz = k;
}
}
len=bfs();
if(len)
printf("Escaped in %d minute(s).\n",len);
else
printf("Trapped!\n");
}
return 0;
}
好吧我再贴一段代码：
/*************************************************************************
> File Name: Dungeon_Master.c
> Author: zhanghaoran
> Mail: 467908670@qq.com
> Created Time: 2015年06月01日 星期一 18时22分17秒
************************************************************************/
#include <stdio.h>
#include <string.h>
struct q{
int x;
int y;
int z;
}q[30000];
int length[30000];
int dx[] = {1, -1, 0, 0, 0, 0};
int dy[] = {0, 0, 0, 0, -1, 1};
int dz[] = {0, 0, -1, 1, 0, 0};
int flag[40][40][40];
char map[40][40][40];
int L, R, C, nx, ny, nz;
int bfs(){
int rear;
int front;
int xx, yy, zz;
int i;
memset(flag, 0, sizeof(flag));
memset(length, 0, sizeof(length));
q[0].x = nx;
q[0].y = ny;
q[0].z = nz;
front = rear = 0;
while(front <= rear){
for(i = 0; i < 6; i ++){
xx = q[front].x + dx[i];
yy = q[front].y + dy[i];
zz = q[front].z + dz[i];
if(!flag[xx][yy][zz] && (map[xx][yy][zz] == '.' || map[xx][yy][zz] == 'E') && xx >= 0 && xx < L && yy < R && yy >= 0 && zz >= 0 && zz < C){
flag[xx][yy][zz] = 1;
q[++rear].x = xx;
q[rear].y = yy;
q[rear].z = zz;
length[rear] = length[front] + 1;
if(map[xx][yy][zz] == 'E')
return length[rear];
}
}
front ++;
}
return 0;
}
int main(void){
int i, j, k, len;
while(scanf("%d%d%d\n" ,&L, &R, &C)){
if(L == 0 && R == 0 && C == 0)
break;
for(i = 0; i < L; i ++, getchar())
for(j = 0; j < R; j ++, getchar())
for(k = 0; k < C; k ++){
scanf("%c", &map[i][j][k]);
if(map[i][j][k] == 'S'){
nx = i;
ny = j;
nz = k;
}
}
len = bfs();
if(len)
printf("Escaped in %d minute(s).\n", len);
else printf("Trapped\n");
}
return 0;
}这两个代码是不是看起来完全一样？我也是这么认为的。。。然而我并不知道为什么第一个提交的时候就是AC，第二个就是WA，第一个是因为我第二个找不到错的地方重新写了一遍，这个问题有待解决。这就是为什么说这个题简直了。。
