藍橋杯-迷宮(BFS+DFS)

0.題目

1.題解

1.1 BFS搜尋 + DFS輸出路徑

思路

主要跟走迷宮那題不一樣的地方在於需要輸出路徑,這裡如何輸出路徑呢?
我們對於每一個節點,均記錄其父節點,之後使用DFS反向遞迴回(0,0)輸出節點資訊,然後逐漸回溯到終點即可.

對於步數相同的,按位元組序來選,我們可以透過安排 D L R U 的順序即可,位元組序小的組合必定會被優先嚐試,嘗試失敗才會嘗試位元組序更大的組合
int dir[4][2]={{1,0},{0,-1},{0,1},{-1,0}};//D L R U

程式碼

#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];//標記
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;//儲存D L R U
};

node father[maxn][maxn];//當前節點的父節點
node now,nex;//指向當前和下一個位置

void dfs(int x,int y)//遞迴列印
{
    if(x==0&&y==0)//找到起點開始正向列印路徑
        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個方向
        {
            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')//判斷是否超出範圍，是否用過，是否為1
            {
                vis[tx][ty]=true;//標記為用過

                nex.x=tx;
                nex.y=ty;
                nex.d=now.d+1;
                q.push(nex);//壓入佇列

                father[tx][ty].x=now.x;//儲存父節點座標
                father[tx][ty].y=now.y;
                if(i==0)//儲存路徑
                    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);//列印路徑

    return 0;
}