蓝桥杯-迷宫(BFS+DFS)

0.题目

1.题解

1.1 BFS搜索 + DFS输出路径

思路

主要跟走迷宫那题不一样的地方在于需要输出路径,这里如何输出路径呢?
我们对于每一个节点,均记录其父节点,之后使用DFS反向递归回(0,0)输出节点信息,然后逐渐回溯到终点即可.

对于步数相同的,按字节序来选,我们可以通过安排 D L R U 的顺序即可,字节序小的组合必定会被优先尝试,尝试失败才会尝试字节序更大的组合
int dir[4][2]={{1,0},{0,-1},{0,1},{-1,0}};//D L R U

代码

#include<bits/stdc++.h>
using namespace std;
#define maxn 2000

string maze[maxn]= {
                  "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];//标记
int dir[4][2]={{1,0},{0,-1},{0,1},{-1,0}};//D L R U

bool in(int x,int y)
{
    return x<30&&x>=0&&y>=0&&y<50;
}

struct node
{
    int x,y,d;
    char pos;//存储D L R U
};

node father[maxn][maxn];//当前节点的父节点
node now,nex;//指向当前和下一个位置

void dfs(int x,int y)//递归打印
{
    if(x==0&&y==0)//找到起点开始正向打印路径
        return;
    else
        dfs(father[x][y].x,father[x][y].y);

    cout<<father[x][y].pos;
}

void bfs(int x,int y)
{
    queue<node> q;

    now.x=x;
    now.y=y;
    now.d=0;
    q.push(now);

    vis[x][y]=true;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(int i=0;i<4;i++)//走下左右上按字典序的四个方向
        {
            int tx=now.x+dir[i][0];
            int ty=now.y+dir[i][1];
            if(in(tx,ty)&&!vis[tx][ty]&&maze[tx][ty]!='1')//判断是否超出范围，是否用过，是否为1
            {
                vis[tx][ty]=true;//标记为用过

                nex.x=tx;
                nex.y=ty;
                nex.d=now.d+1;
                q.push(nex);//压入队列

                father[tx][ty].x=now.x;//存储父节点坐标
                father[tx][ty].y=now.y;
                if(i==0)//存储路径
                    father[tx][ty].pos='D';
                else if(i==1)
                    father[tx][ty].pos='L';
                else if(i==2)
                    father[tx][ty].pos='R';
                else if(i==3)
                    father[tx][ty].pos='U';


            }
        }
    }
}

int main()
{

    bfs(0,0);
    dfs(29,49);//打印路径

    return 0;
}