这道题我们得知道象棋里的马能怎么走,应该是按日字走的

如图，就这么走，我们按基础的bfs遍历一下标记每个格子的最短的路径就行了，没到达的格子就设置为-1

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int N = 410;
int n, m, x, y;
int dict[N][N];
int dx[] = { -2,-1,1,2,2,1,-1,-2 };
int dy[] = { 1,2,2,1,-1,-2,-2,-1 };
typedef pair<int, int> PII;
void bfs() {memset(dict, -1, sizeof(dict));queue<PII> q;q.push({ x,y });dict[x][y] = 0;while (q.size()){auto t = q.front();q.pop();int x = t.first;int y = t.second;for (int k = 0; k < 8; k++){int px = x + dx[k];int py = y + dy[k];if (px<1 || py<1 || px>n || py>m)continue;if (dict[px][py] != -1) continue;dict[px][py] = dict[x][y] + 1;q.push({ px,py });}}}
int main()
{cin >> n >> m >> x >> y;bfs();for (int i = 1; i <= n; i++){for (int j = 1; j <= m; j++){cout << dict[i][j] << " ";}cout << endl;}return 0;
}