深搜和广搜
Dungeon Master
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 58498 | Accepted: 21525 |
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
深搜只是确定它是否存在；（深搜时间复杂度远远大于广搜，所以一般找到结果之后都会选择直接退出）
广搜才是确定它的最短路径；
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int vis[36][36][36];
int ans;
char tu[36][36][36];
int l,r,c;
int flag;
int cnt;
int zhong;
int movee[19]= {0, 0,1,0, 0,-1,0, 0,0,1, 0,0,-1, 1,0,0, -1,0,0 };
//1 (1) 4(2) 7(3) 10(4) 13(5) 16(6)
void in(int high,int x,int y)
{
if(tu[high][x][y]=='E' && cnt < zhong)
{
zhong=cnt;
ans=cnt;
return;
}
else
{
for(int i=1; i<=16; i=i+3)
{
high=high+movee[i];
x=x+movee[i+1];
y=y+movee[i+2];
if( ( tu[high][x][y]=='.' || tu[high][x][y]=='E' ) && vis[high][x][y]==0 && high>=1 && high<=l && x>=1 && x<=r &&y>=1 &&y<=c )
{
vis[high][x][y]=1;
cnt++;
in(high,x,y);
cnt--;
vis[high][x][y]=0;
}
high=high-movee[i];
x=x-movee[i+1];
y=y-movee[i+2];
}
}
return;
}
int main()
{
while((cin>>l>>r>>c) && l && r &&c)
{
memset(vis,0,sizeof(vis));
cnt=0;
int high,x,y;
int high_end,x_end,y_end;
ans=0;
zhong=0x3f3f3f;
for(int i=1; i<=l; i++)
{
for(int j=1; j<=r; j++)
{
for(int k=1; k<=c; k++)
{
cin>>tu[i][j][k];
if( tu[i][j][k]=='S' )
{
high=i;
x=j;
y=k;
//cout<<i<<j<<k<<endl;
}
if( tu[i][j][k]=='E' )
{
high_end=i;
x_end=j;
y_end=k;
}
}
}
}
vis[high][x][y]=1;
in(high,x,y);
if(ans)
{
printf("Escaped in %d minute(s).\n",ans);
}
else
{
printf("Trapped!\n");
}
}
}
#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
char map[30][30][30]; //记录节点信息
int vis[30][30][30]; //标记是否访问
int base[6][3] = { {-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1} };
int L, R, C;
typedef struct
{
int x, y, z; //位置坐标
int step; //出发点到该点的步数
} node;
node s; //起点
node e; //终点
node curp; //跳出循环时的节点
bool success(node cur)
{
if (cur.x == e.x && cur.y == e.y && cur.z == e.z)
return true;
else
return false;
}
bool check(int x, int y, int z)
{
if ( (x >= 0) && (x < L) && (y >= 0) && (y < R) && (z >= 0) && (z < C) && (!vis[x][y][z]) && (map[x][y][z] == '.' || map[x][y][z] == 'E'))
return true;
else
return false;
}
void bfs()
{
queue<node> q;
q.push(s);
while (!q.empty())
{
curp = q.front();
q.pop();
if ( success(curp) )
return;
else
{
vis[curp.x][curp.y][curp.z] = 1;
node next;
for (int i = 0; i < 6; i++)
{
next.x = curp.x + base[i][0];
next.y = curp.y + base[i][1];
next.z = curp.z + base[i][2];
if ( check(next.x, next.y, next.z) ) //扩展队列
{
next.step = curp.step + 1;
vis[next.x][next.y][next.z] = 1;
q.push(next);
}
}
}
}
}
int main()
{
while (scanf("%d%d%d", &L, &R, &C))
{
if((L == 0) && (R == 0) && (C == 0))
break;
memset(vis, 0, sizeof(vis));
for (int i = 0; i < L; i++)
{
for (int j = 0; j < R; j++)
{
for (int k = 0; k < C; k++)
{
cin>>map[i][j][k];
if (map[i][j][k] == 'S')
{
s.x = i;
s.y = j;
s.z = k;
s.step = 0;
}
else if (map[i][j][k] == 'E')
{
e.x = i;
e.y = j;
e.z = k;
}
}
}
}
bfs();
if ( curp.x == e.x && curp.y == e.y && curp.z == e.z )
printf("Escaped in %d minute(s).\n", curp.step);
else
printf("Trapped!\n");
}
return 0;
}