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SWJTU 2184迷宫

时间:2014-12-11 22:16:24      阅读:265      评论:0      收藏:0      [点我收藏+]

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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

bubuko.com,布布扣

Source

 

水题,记录开始到终点移动每一步的方向

 

bubuko.com,布布扣
 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

 

SWJTU 2184迷宫

标签:des   style   blog   http   io   ar   color   os   sp   

原文地址:http://www.cnblogs.com/usedrosee/p/4158542.html

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