Problem Description
The following figure shows a plan view of a maze, where the obstacles marked 1 are obstacles, and the places marked 0 are passable places.
010000
000100
001001
110000
The entrance of the maze is the upper left corner, and the exit is the lower right corner. In the maze, you can only walk from one position to one of its four directions, up, down, left, and right.
For the maze above, starting from the entrance, you can pass through the maze in the order of DRRURRDDDR, a total of 10 steps.
Among them, D, U, L, and R indicate go down, up, left, and right respectively.
For this more complicated maze (30 rows and 50 columns)
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
Please find a way through the maze that uses the least number of steps. Under the premise of the least number of steps, please find the one with the least lexicographical order as the answer.
Please note that D <L <R <U in lexicographic order.
Answer submission
This is a question that fills in the blanks with the result. You only need to calculate the result and submit it.
The result of this question is a string containing the four letters D, U, L, and R. When submitting the answer, only fill in this string, and fill in the extra content will not be scored.
answer
For the shortest path problem of the maze class, use bfs to find it. Note that if you use dfs, you will not get the result for a long time.
In the question, if the number of steps is the same, the output with the smallest lexicographic order is required. Then we can traverse the four directions in the order of DLRU, which ensures that the first solution found satisfies both the smallest number of steps and the smallest lexicographical order.
code show as below
#include<bits/stdc++.h>
using namespace std;
struct node //结构体node记录xy坐标和走的顺序
{
int x,y;string s;
node(int xx,int yy,string ss)
{
x=xx;y=yy,s=ss;
}
};
string mp[100];
int dir[4][2]={
{
1,0},{
0,-1},{
0,1},{
-1,0}};
char zimu[4]={
'D','L','R','U'};
bool visited[100][100];
int m=30,n=50;
queue<node > qn;
void bfs(int x,int y)
{
qn.push(node(0,0,""));
visited[x][y]=1;
while(!qn.empty())
{
node t=qn.front();
qn.pop();
for (int i = 0; i < 4; i++)
{
int tx=t.x+dir[i][0];
int ty=t.y+dir[i][1];
if(tx>=0&&tx<m&&ty>=0&&ty<n&&!visited[tx][ty]&&mp[tx][ty]!='1')
{
if(tx==m-1&&ty==n-1)
{
cout<<t.s<<zimu[i]<<endl; //注意也要将zimu[i]一并输出因为此时它还没有被加到t.s
return;
}
else
{
visited[tx][ty]=1;
qn.push(node(tx,ty,t.s+zimu[i]));
}
}
}
}
}
int main()
{
for (int i = 0; i < m; i++)
{
cin>>mp[i];
}
bfs(0,0);
return 0;
}
The result is:
DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRRRRRDDRRDDLLLDDRRDDRRRRRRRRRDDDDDLLDDLLLDLDDDLDDRRRRRRRRRDD
bfs template
queue<type> q;
q.push(初始状态);
while (!q.empty())
{
type t = q.front() ;
q.pop();
遍历 t 的各个Next状态 next
{
if (next is legal)
q.push(next的状态); 计数或维护等;
}
}