LQ0087 迷宫BFS

Posted 2022-10-24 海岛Blog

题目来源：蓝桥杯2019初赛 C++ A组D题

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。

输入格式
见文件：maze.txt

问题分析
求最短路径，用BFS来实现。
该题算是一个模板题。

AC的C++语言程序如下：

/* LQ0087 迷宫 */

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>

using namespace std;

#define N 30
#define M 50

char g[][M + 1] = 
    "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",
;

int drow[] = 1, 0, 0, -1;
int dcol[] = 0, -1, 1, 0;
char dir[] = "DLRU";
bool vis[N][M];

struct Node 
    int row, col;
    string p;
    Node(int row, int col, string p):row(row), col(col), p(p)
;

void bfs()

    memset(vis, false, sizeof vis);
    vis[0][0] = true;

    queue<Node> q;
    q.push(Node(0, 0, ""));
    while (!q.empty()) 
        Node t = q.front();
        q.pop();

        if (t.row == N - 1 && t.col == M - 1) 
            cout << t.p << endl;
            break;
         else 
            for (int i = 0; i < 4; i++) 
                int nextrow = t.row + drow[i];
                int nextcol = t.col + dcol[i];
                if (0 <= nextrow && nextrow < N && 0 <= nextcol && nextcol <= M &&
                        !vis[nextrow][nextcol] && g[nextrow][nextcol] != '1') 
                    Node node(nextrow, nextcol, t.p + dir[i]);
                    q.push(node);
                    vis[nextrow][nextcol] = true;
                
            
        
    


int main()

    bfs();

    return 0;