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(jiǎn)單的一個(gè)搜索題目，廣搜和深搜都要用
題目意思，有一個(gè)迷宮，現(xiàn)在想知道總是沿著迷宮左側(cè)墻壁走的能走多少步，沿著迷宮右側(cè)墻壁走的能走多少步
最少走多少步能出去
我覺(jué)著這個(gè)題目關(guān)鍵在于如何處理怎樣沿著左右墻壁走
我一開(kāi)始以為設(shè)好總是右拐左拐的方向就行了，結(jié)果樣例就不過(guò)，最后輸出一看，有個(gè)地方一直來(lái)回走了，死循環(huán)了
這里怎么處理呢？
我也不會(huì)，網(wǎng)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(lái)寫(xiě)思路
除了這種方法，還沒(méi)想到別的方法，額，糾結(jié)了
1
#include<stdio.h>
2#include<math.h>
3#include<string.h>
4#define MAX 45
5#define MX 16000
6struct node
7
{
8
int x,y,d;
9
};
10int w,h,sx,sy,ans1,ans2,ans3;
11int dx[4][2]={{-1,0},{0,-1},{1,0},{0,1}};
12int q1;
13char map[MAX][MAX];
14short mark[MAX][MAX];
15int inin(int xx,int yy)
16{
17 return xx>=0&&xx<h&&yy>=0&&yy<w;
18}
19void 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}
41int 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}
58int 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}
73int 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}
102void work()
103{
104 int i,j;
105 ans3=bfs();
106 ans1=dfs1(sx,sy,q1);
107 ans2=dfs2(sx,sy,q1);
108}
109int 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