Children of the Candy Corn
Time Limit: 1000MS |
|
Memory Limit: 65536K |
Total Submissions: 6055 |
|
Accepted: 2640 |
Description
The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and must wander through the maze facing zombies, chainsaw-wielding psychopaths, hippies, and other terrors on their quest to find the exit.
One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.)
As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.
Input
Input to this problem will begin with a line containing a single integer n indicating the number of mazes. Each maze will consist of one line with a width, w, and height, h (3 <= w, h <= 40), followed by h lines of w characters each that represent the maze layout. Walls are represented by hash marks ('#'), empty space by periods ('.'), the start by an 'S' and the exit by an 'E'.
Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#').
You may assume that the maze exit is always reachable from the start point.
Output
For each maze in the input, output on a single line the number of (not necessarily unique) squares that a person would visit (including the 'S' and 'E') for (in order) the left, right, and shortest paths, separated by a single space each. Movement from one square to another is only allowed in the horizontal or vertical direction; movement along the diagonals is not allowed.
Sample Input
2
8 8
########
#......#
#.####.#
#.####.#
#.####.#
#.####.#
#...#..#
#S#E####
9 5
#########
#.#.#.#.#
S.......E
#.#.#.#.#
#########
Sample Output
37 5 5
17 17 9
挺簡單的一個搜索題目,廣搜和深搜都要用
題目意思,有一個迷宮,現在想知道總是沿著迷宮左側墻壁走的能走多少步,沿著迷宮右側墻壁走的能走多少步
最少走多少步能出去
我覺著這個題目關鍵在于如何處理怎樣沿著左右墻壁走
我一開始以為設好總是右拐左拐的方向就行了,結果樣例就不過,最后輸出一看,有個地方一直來回走了,死循環了
這里怎么處理呢?
我也不會,網上看到神一般的代碼
for (i=xx+1;i>=xx-2;i--)
{
j=(8+i)&3;//????
}
真心不明白啥意思,等弄懂了再回來寫思路
除了這種方法,還沒想到別的方法,額,糾結了
1
#include<stdio.h>
2
#include<math.h>
3
#include<string.h>
4
#define MAX 45
5
#define MX 16000
6
struct node
7

{
8
int x,y,d;
9
};
10
int w,h,sx,sy,ans1,ans2,ans3;
11
int dx[4][2]=
{
{-1,0},
{0,-1},
{1,0},
{0,1}};
12
int q1;
13
char map[MAX][MAX];
14
short mark[MAX][MAX];
15
int inin(int xx,int yy)
16

{
17
return xx>=0&&xx<h&&yy>=0&&yy<w;
18
}
19
void init()
20

{
21
int i,j;
22
int flag;
23
flag=0;
24
scanf("%d%d",&w,&h);
25
for (i=0;i<h;i++)
26
{
27
scanf("%s",&map[i]);
28
for (j=0;j<w;j++)
29
if (map[i][j]=='S'&&!flag)
30
{
31
sx=i;sy=j;
32
flag=1;
33
}
34
}
35
for (i=0;i<4;i++)
36
if (inin(sx+dx[i][0],sy+dx[i][1])&&map[sx+dx[i][0]][sy+dx[i][1]]=='.')
37
{
38
q1=i;break;
39
}
40
}
41
int dfs1(int x,int y,int xx)
42

{
43
int j,i,nx,ny;
44
if (map[x][y]=='E')
45
{
46
return 1;
47
}
48
for (i=xx+1;i>=xx-2;i--)
49
{
50
j=(8+i)&3;//????
51
nx=x+dx[j][0];ny=y+dx[j][1];
52
if ((inin(nx,ny))&&(map[nx][ny]!='#'))
53
{
54
return 1+dfs1(nx,ny,j);
55
}
56
}
57
}
58
int dfs2(int x,int y,int xx)
59

{
60
int j,i,nx,ny;
61
if (map[x][y]=='E')
62
return 1;
63
for (i=xx-1;i<=xx+2;i++)
64
{
65
j=(8+i)&3;//????
66
nx=x+dx[j][0];ny=y+dx[j][1];
67
if ((inin(nx,ny))&&(map[nx][ny]!='#'))
68
{
69
return 1+dfs2(nx,ny,j);
70
}
71
}
72
}
73
int bfs()
74

{
75
int i,j,head,tail,nx,ny;
76
struct node q[MX],now;
77
memset(mark,0,sizeof(mark));
78
head=0;tail=1;
79
mark[sx][sy]=1;
80
q[tail].x=sx;q[tail].y=sy;q[tail].d=1;
81
while (head<tail)
82
{
83
head++;
84
now=q[head];
85
for (i=0;i<4;i++)
86
{
87
nx=now.x+dx[i][0];
88
ny=now.y+dx[i][1];
89
if (inin(nx,ny)&&map[nx][ny]!='#'&&!mark[nx][ny])
90
{
91
if (map[nx][ny]=='E')
92
{
93
return now.d+1;
94
}
95
tail++;
96
q[tail].x=nx;q[tail].y=ny;q[tail].d=now.d+1;
97
mark[nx][ny]=1;
98
}
99
}
100
}
101
}
102
void work()
103

{
104
int i,j;
105
ans3=bfs();
106
ans1=dfs1(sx,sy,q1);
107
ans2=dfs2(sx,sy,q1);
108
}
109
int main()
110

{
111
int t,ii;
112
scanf("%d",&t);
113
for (ii=1;ii<=t;ii++)
114
{
115
memset(map,0,sizeof(map));
116
init();
117
work();
118
printf("%d %d %d\n",ans1,ans2,ans3);
119
}
120
return 0;
121
}
122