90分,1个TLE
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90分,1个TLE
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zhou_yu_cheng楼主2025/8/4 22:46
#include <bits/stdc++.h>
using namespace std;
const int N = 410; // 棋盘最大尺寸
int dist[N][N]; // 记录从起点到每个点的最短路径
int n, m; // 棋盘大小
int dx[] = {-2, -2, -1, 1, 2, 2, 1, -1}; // 马的移动方向
int dy[] = {-1, 1, 2, 2, 1, -1, -2, -2};
void dfs(int x, int y, int steps) {
	dist[x][y] = steps; // 更新当前点的步数
	for (int i = 0; i < 8; i++) {
		int nx = x + dx[i], ny = y + dy[i];
		if (nx >= 1 && nx <= n && ny >= 1 && ny <= m && dist[nx][ny] > steps + 1) {
			dfs(nx, ny, steps + 1); // 深度递归
		}
	}
}
int main() {
	memset(dist, 0x3f, sizeof(dist)); // 初始化为无穷大
	cin >> n >> m;
	int startX, startY;
	cin >> startX >> startY;
	dfs(startX, startY, 0); // 从起点开始DFS
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			if (dist[i][j] != 0x3f3f3f3f) cout << dist[i][j] << " ";
			else cout << -1 << " "; // 无法到达的点输出-1
		}
		cout << endl;
	}
	return 0;
}//马的遍历
2025/8/4 22:46
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