洛谷P1443 马的遍历(bfs)

https://www.luogu.org/problem/P1443

思路：马走日字格，它可以走十六个方向,int dir[8][2]= {2,1,1,2,-2,1,-1,2,2,-1,1,-2,-2,-1,-1,-2}; 将马可以走到的每一步入队列，记录步数

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <cstring>
#include <map>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
typedef long long lll;
const int maxn = 200005;
const lll INF = 0x3f3f3f3f3f;
int dir[8][2]= {2,1,1,2,-2,1,-1,2,2,-1,1,-2,-2,-1,-1,-2};
int dir2[4][2]= {0,1,0,-1,1,0,-1,0};
bool flag;
int a[1005][1005],b[1005][1005],n,m,sx,sy;
string s[1005];
struct node
{
    int x,y;
    node() {};
    node(int xx,int yy):x(xx),y(yy) {};
};

void bfs(int x,int y)
{
    //  cout << 1 <<endl;
    queue<node>q;
    q.push(node(x,y));
//    b[x][y]=1;
    a[x][y] = 0;
    while(!q.empty())
    {
        node p = q.front();
        q.pop();
        for(int i=0; i<8; i++)
        {
            int fx = p.x + dir[i][0],fy = p.y + dir[i][1];
            if(fx < 1 || fx > n || fy < 1 || fy > m ||a[fx][fy]!=-1||(fx == sx&&fy == sy)) continue;
            q.push(node(fx,fy));
            a[fx][fy] = a[p.x][p.y]+1;
           // b[fx][fy]=1;
        }
    }
}
int main()
{
    memset(a,-1,sizeof(a));
    scanf("%d%d%d%d",&n,&m,&sx,&sy);
    bfs(sx,sy);
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=m; j++)
            printf("%-5d",a[i][j]);
        printf("
");
    }
    return 0;
}