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#include <bits/stdc++.h>
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using namespace std;
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int n, m; //n*m的棋盘
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int x, y; //马的位置
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const int N = 410;
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int ans[N][N]; //到任意点需要最少走几步
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//坐标
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struct coord {
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int x, y;
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};
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queue<coord> q;
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//马的八大方向坐标位移
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int walk[8][2] = {{2, 1},
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{1, 2},
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{-1, 2},
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{-2, 1},
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{-2, -1},
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{-1, -2},
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{1, -2},
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{2, -1}};
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int main() {
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//输入棋盘大小和马的位置
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cin >> n >> m >> x >> y;
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//标识为-1,表示没有探索过
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memset(ans, -1, sizeof ans);
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//马入队列
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q.push({x, y});
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ans[x][y] = 0;//出发点标识为最少走0步
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while (!q.empty()) {
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coord u = q.front();
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q.pop();
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for (int k = 0; k < 8; k++) {//8个方向
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int x = u.x + walk[k][0], y = u.y + walk[k][1];//可能达到的下一步坐标
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if (x < 1 || x > n || y < 1 || y > m || ans[x][y] != -1)continue;
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//步数等于上一级来的位置+1
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ans[x][y] = ans[u.x][u.y] + 1;
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//入队列
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q.push({x, y});
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}
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}
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//输出结果
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for (int i = 1; i <= n; i++) {
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for (int j = 1; j <= m; j++)
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printf("%-5d", ans[i][j]); //左对齐
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puts("");
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}
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return 0;
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}
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