2019年第十届C/C++ A组蓝桥杯省赛第四题：迷宫

Posted 2023-02-14 梦西空

题目描述
下图给出了一个迷宫的平面图，其中标记为 1 的为障碍，标记为 0 的为可 以通行的地方。

010000

000100

001001

110000

迷宫的入口为左上角，出口为右下角，在迷宫中，只能从一个位置走到这 个它的上、下、左、右四个方向之一。 对于上面的迷宫，从入口开始，可以按DRRURRDDDR 的顺序通过迷宫， 一共 10 步。其中 D、U、L、R 分别表示向下、向上、向左、向右走。 对于下面这个更复杂的迷宫（30 行 50 列），请找出一种通过迷宫的方式， 其使用的步数最少，在步数最少的前提下，请找出字典序最小的一个作为答案。 请注意在字典序中D<L<R<U。

01010101001011001001010110010110100100001000101010
00001000100000101010010000100000001001100110100101
01111011010010001000001101001011100011000000010000
01000000001010100011010000101000001010101011001011
00011111000000101000010010100010100000101100000000
11001000110101000010101100011010011010101011110111
00011011010101001001001010000001000101001110000000
10100000101000100110101010111110011000010000111010
00111000001010100001100010000001000101001100001001
11000110100001110010001001010101010101010001101000
00010000100100000101001010101110100010101010000101
11100100101001001000010000010101010100100100010100
00000010000000101011001111010001100000101010100011
10101010011100001000011000010110011110110100001000
10101010100001101010100101000010100000111011101001
10000000101100010000101100101101001011100000000100
10101001000000010100100001000100000100011110101001
00101001010101101001010100011010101101110000110101
11001010000100001100000010100101000001000111000010
00001000110000110101101000000100101001001000011101
10100101000101000000001110110010110101101010100001
00101000010000110101010000100010001001000100010101
10100001000110010001000010101001010101011111010010
00000100101000000110010100101001000001000000000010
11010000001001110111001001000011101001011011101000
00000110100010001000100000001000011101000000110011
10101000101000100010001111100010101001010000001000
10000010100101001010110000000100101010001011101000
00111100001000010000000110111000000001000000001011
10000001100111010111010001000110111010101101111000

/**
 * @Time    :2021-04-14-19.33
 * @Author  :Yang JiaBin
 * @version :V 1.0
 * @desc    :给出一个迷宫的地图，其中为障碍，为可通行的地方。
 *迷宫的入口为左上角，出口为右下角，在迷宫中，(0,1)
 *只能从一个位置走到这 个它的上、下、左、右四个方向之一。
 *用 D、U、L、R 分别表示向下、向上、向左、向右走，
 *请记录你的最短行径路径，并输出其中字典序最小的一个。
 * bfs解决迷宫最短路问题
 * 利用BFS算法给出的路径必然是一条最短路径
 */
#include<bits/stdc++.h>
using namespace std;
const int maxn=105;
//按DLRU顺序可得字典序最小
char str[]='D','L','R','U';
int fw[][2]=1,0,0,-1,0,1,-1,0;
int n=30;//迷宫的行
int m=50;//迷宫的列
string grp[]=
"01010101001011001001010110010110100100001000101010",
"00001000100000101010010000100000001001100110100101",
"01111011010010001000001101001011100011000000010000",
"01000000001010100011010000101000001010101011001011",
"00011111000000101000010010100010100000101100000000",
"11001000110101000010101100011010011010101011110111",
"00011011010101001001001010000001000101001110000000",
"10100000101000100110101010111110011000010000111010",
"00111000001010100001100010000001000101001100001001",
"11000110100001110010001001010101010101010001101000",
"00010000100100000101001010101110100010101010000101",
"11100100101001001000010000010101010100100100010100",
"00000010000000101011001111010001100000101010100011",
"10101010011100001000011000010110011110110100001000",
"10101010100001101010100101000010100000111011101001",
"10000000101100010000101100101101001011100000000100",
"10101001000000010100100001000100000100011110101001",
"00101001010101101001010100011010101101110000110101",
"11001010000100001100000010100101000001000111000010",
"00001000110000110101101000000100101001001000011101",
"10100101000101000000001110110010110101101010100001",
"00101000010000110101010000100010001001000100010101",
"10100001000110010001000010101001010101011111010010",
"00000100101000000110010100101001000001000000000010",
"11010000001001110111001001000011101001011011101000",
"00000110100010001000100000001000011101000000110011",
"10101000101000100010001111100010101001010000001000",
"10000010100101001010110000000100101010001011101000",
"00111100001000010000000110111000000001000000001011",
"10000001100111010111010001000110111010101101111000";

bool vis[maxn][maxn];

struct N

     int x;
     int y;
     string road;
;
N from[maxn][maxn];//保存路径
char fromfw[maxn][maxn];//访问方位路径

deque<N> que;

bool judge(int x,int y)

     if(x<0||x>n-1||y<0||y>m-1)
          return false;
     
     if(grp[x][y]=='1')
          return false;
     
     return true;


bool bfs(int x0,int y0,int x1,int y1)

     que.push_back((N)x0,y0,"");
     vis[x0][y0]=true;

     while(!que.empty())
          N t=que.front();
          que.pop_front();
          if(t.x==n-1&&t.y==m-1) 

			cout<<t.road<<endl;   //找到了路径
			break;
		
          for(int i=0;i<4;i++)
               N s;
               s.x=t.x+fw[i][0];
               s.y=t.y+fw[i][1];
               if(judge(s.x,s.y))
                    if(!vis[s.x][s.y])
                         vis[s.x][s.y]=true;
                         s.road=t.road+str[i];
                         que.push_back(s);

                    
               


          

     

     return false;



int main()

     bfs(0,0,n-1,m-1);
	return 0;