SWJTU 2184迷宫

DongGua: Save our SWJTU!

Time Limit:2000MS  Memory Limit:65536K
Total Submit:46 Accepted:22

Description

One night ,Desgard_Duan was caught by an alien named “DG”when he was sleeping soundly.
“Now,you are in a extremely complicated maze, if you can’t escape from this maze,you will be trapped in it all your life!And I will destroy SWJTU OJ! hhhhhhhhhhhhhhh”
“Why you choose me?!”
“OK,shut up,now I will give you some prompts.This is a N rows and M columns maze ,and then your start point (X1,Y1),the end of maze (X2,Y2).For every location,i will give you a value which means the value of walls around it.”
“What’s the value mean?”(mad)
“Xiao Xiao Duan Donggua,so clever ya!
If the location has a wall above it the variable has 8 added to it.
If the location has a wall to the left of it the variable has 4 added to it.
If the location has a wall below it the variable has 2 added to it.
If the location has a wall to the right of it the variable has 1 added to it.”
But,as you know ,Duan Donggua’s ID is not so reliable.Can you help him?

Input

For each case, the first line of input contains two numbers: m and n (0 <= m, n <= 600).
The second line contains two number X1, Y1. They are start point.
As the second line, two number X2, Y2 are the end of maze. (0 <= X1, X2 < m, 0 <= Y1, Y2 < n)
The following m lines include n numbers each. And this is the Matrix to describe the maze.

Output

Your program should output a text file with two lines:
The first line should be the number of moves required to get from the start location to the end location.
The second line should be a list of moves that will trace a path from the start position to the end position without going through any walls. Each move is represented by a single letter U, D, L or R for Up, Down, Left or Right respectively.

Sample Input

 

5 4

2 2

4 0

14 9 12 9

14 2 3 5

13 12 8 3

5 5 6 9

6 2 11 7

 

Sample Output

 

4

LDDL

 

Hint

Source

 

水题，记录开始到终点移动每一步的方向

 

 1 #include "iostream"

 2 #include "string"

 3 #include "memory.h"

 4 #include "queue"

 5 #include "algorithm"

 6 #include "cstdio"

 7 using namespace std;

 8 #define MAXN 1111

 9 const char  dir[] = {'U','D','L','R'};

10 const int dx[] = {-1,1,0,0},dy[] = {0,0,-1,1};

11 int map[MAXN][MAXN];

12 int vis[MAXN][MAXN];

13 string wall[MAXN][MAXN];

14 int n,m;

15 int sx,sy,fx,fy;

16 

17 struct node {

18     int x,y,time;

19     string path;

20 }now;

21 

22 

23 void gao(int temp,int i,int j)

24 {

25     wall[i][j] = "";

26     if (temp - 8 >= 0) temp -= 8, wall[i][j] += 'U';

27     if (temp - 4 >= 0) temp -= 4, wall[i][j] += 'L';

28     if (temp - 2 >= 0) temp -= 2, wall[i][j] += 'D';

29     if (temp - 1 >= 0) temp -= 1, wall[i][j] += 'R';

30 }

31 

32 bool judge(int x,int y)

33 {

34     if (x < 0 || y < 0 || x >= m || y >= n) return 1;

35     if (vis[x][y]) return 1;

36     return 0;

37 }

38 void bfs()

39 {

40     queue<node> q;

41     now.x = sx;

42     now.y = sy;

43     now.path ="";

44     now.time = 0;

45     vis[sx][sy] = 1;

46     q.push(now);

47     while (!q.empty())

48     {

49         now = q.front();

50         q.pop();

51         if (now.x == fx && now.y == fy)

52         {

53             cout << now.time << endl;

54             cout << now.path <<endl;

55             return ;

56         }

57         int i;

58         node next;

59         for (i = 0;i < 4; ++ i)

60         {

61             if (wall[now.x][now.y].find(dir[i]) == -1) {

62                 next.x = now.x + dx[i];

63                 next.y = now.y + dy[i];

64             } else continue;

65             if (judge(next.x,next.y)) continue;

66             next.path  = now.path + dir[i];

67             next.time  = now.time + 1;

68             vis[next.x][next.y] = 1;

69             q.push(next);

70         }

71     }

72 

73 }

74 int main()

75 {

76     int i, j;

77     while ( cin >> m >> n ) {

78         memset(vis,0,sizeof(vis));

79         scanf("%d%d%d%d",&sx,&sy,&fx,&fy);

80         for (i = 0;i < m; ++ i)

81             for (j = 0;j < n; ++ j) {

82            scanf("%d",&map[i][j]);

83             gao(map[i][j],i,j);

84         }

85         bfs();

86     }

87     return 0;

88 }

View Code