CF3A Shortest path of the king

CF3A Shortest path of the king

Luogu题地址

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 10;
int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
string s[8] = {"U", "RU", "R", "RD", "D", "LD", "L", "LU"};
vector<string> ans;
struct Node{
    int x, y, step;
}node, S, T;

bool st[N][N];
int d[N][N];
Node q[N];

void bfs()
{
    d[S.x][S.y] = 0;
    queue<Node> q;
    q.push(S);
    st[S.x][S.y] = true;
    while(q.size())
    {
        Node t = q.front();
        q.pop();

        int x = t.x, y = t.y, step = t.step;
        d[x][y] = step;
        for(int i = 0; i < 8; i ++)
        {
            int nx = x + dx[i], ny = y + dy[i];
            if(nx >= 0 && nx <= 7 && ny >= 0 && ny <= 7 && !st[nx][ny])
            {
                node.x = nx, node.y = ny, node.step = step + 1;
                q.push(node);
                st[nx][ny] = true;
            }
        }

    }
}

void find(int x, int y)
{
    for(int i = 0; i < 8; i ++)
    {
        if(x == dx[i] && y == dy[i])
            ans.push_back(s[i]);
    }
}

int main()
{
    char s1[3], s2[3];
    cin >> s1 >> s2;
    S = {s1[1] - '1', s1[0] - 'a', 0};
    T = { s2[1] - '1',s2[0] - 'a', 0};

    bfs();
    cout << d[T.x][T.y] << endl;
    int x = T.x, y = T.y;
    if(d[x][y] == 0) return 0;

    while(1)
    {
        bool flag = false;
        for(int i = 0; i < 8; i ++)
        {
            int nx = x + dx[i], ny = y + dy[i];
            if(nx < 0 || nx > 7 || ny < 0 || ny > 7) continue;
            if(d[nx][ny] == 0)
            {
                find(x - nx, y - ny);
                flag = true;
                break;
            }
            if(d[nx][ny] == d[x][y] - 1)
            {
                find(x - nx, y - ny);
                x = nx, y = ny;
                continue;
            }

        }
        if(flag == true) break;
    }
    int len = ans.size() - 1;
    for(int i = len; i >= 0; i --)
        cout << ans[i] << endl;
}