sdnuoj-1220 之旧地重游

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刚刚A了紫书上的一道BFS题，就正好把其精华应用到了这道题上。

思路

现在想想其实这个输出路径的无非是多了一条输出路径而已，我们知道每个点的前驱都是可以在遍历中确定的，我开了一个三维数组，前两维记录坐标，第三维记录的是从哪个方向到达该点（和紫书的那道题一样）。这样我只需要在BFS中加上记录路径的一句话就行了。最后在输出路径的时候，只需要输出第三维的方向就行了。

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#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string>
#include <cstring>
#include <math.h>
#include <sstream>
#include <queue>
#include <set>
#include <map>
using namespace std;
struct node
{
    int x,y,dir;
};

int dir1[] = {1,0,0,-1};   //向东，向西，向南 向北
int dir2[] = {0,-1,1,0};
string ss = "DLRU", s[55];
node p[55][55][10];
int vis[55][55],n, m;

void print(node u)
{
    vector<node>V;
    while(1)
    {
        V.push_back(u);
        if(u.x == 0 && u.y == 0) break;
        u = p[u.x][u.y][u.dir];
    }
    reverse(V.begin(), V.end());
    cout << V.size() - 1 << '\n';
    for(int i = 1; i < V.size(); ++i)
    {
        cout << ss[V[i].dir];
    }
    cout << '\n';
}

void bfs()
{
    queue<node>Q;
    Q.push({0,0,0});
    vis[0][0] = 1;
    while(!Q.empty())
    {
        node u = Q.front();
        Q.pop();
        if(u.x == n - 1 && u.y == m - 1)
        {
            print(u);
            return ;
        } 
        for(int i = 0; i < 4; ++i)
        {
            node v;
            v.dir = i;
            v.x = u.x + dir1[i];
            v.y = u.y + dir2[i];
            if(v.x < 0 || v.x >= n || v.y < 0 || v.y >= m || vis[v.x][v.y] || s[v.x][v.y] == '1')
                continue;
            p[v.x][v.y][i] = u;        //记录路径 这一句话就够了
            vis[v.x][v.y] = 1;
            Q.push(v);
        }
    }
}


int main()
{
    while(scanf("%d %d",&n,&m) != EOF)
    {
        memset(vis, 0,sizeof(vis));
        for(int i = 0; i < n; ++i)
        {
            cin >> s[i];
        }
        bfs();
    }
}

最后附上2019蓝桥杯B组迷宫题数据（30行50列），和这题一样，输出路径

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