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原创最新推荐文章于 2026-08-16 10:41:13 发布 210 阅读 AI 写同款· GEO 优化 ›

【信息科学与工程学】计算机科学与自动化——第十篇芯片设计16 指令集12 编号类型领域模型配方问题【层次性+分级分类+关联性/依赖性+问题详细定义】问题的详细数学分析及算法分析（含逐步推理方程式）参数列表及参数的数值设计关联知识和法律法规2863厂商-架构-指令CPU / ARM AArch64ARM · SHA3 (SHA3 旋转加异或)层次：EL0 → NEON 64位旋转加异或（dest=a⊕(b≪1)）。分级：计算 v0=v1⊕(v2≪1)（64位元素）。依赖：ARMv8.2-A SHA3。问题定义：向量化旋转加异或，用于SHA-3 θ步。逐阅读详情

#include <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define IMAX 2147483646
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define c unsigned long long
#define uint unsigned int

using namespace std;

const int dx[] = { 1,1,-1,-1,2,2,-2,-2 };
const int dy[] = { 2,-2,2,-2,1,-1,1,-1 };
struct Node {
	int x, y, val;
}a,te;
queue<Node> q;
char mp[155][155];
int n, m;
int v[155][155], sx, sy, tx, ty,ans;

int main() {
	scanf("%d%d", &m, &n);
	for (int i = 0; i < n; i++) {
		scanf("%s", mp[i]);
		for (int j = 0; j < m; j++)
			if (mp[i][j] == 'K')sx = i, sy = j;
	}
	for (int i = 0; i < n; i++)
		cout << mp[i] << endl;
	cout << sx << " " << sy << endl;
	a.x = sx, a.y = sy, a.val = 0;
	q.push(a);
	ans = -1;
	while (!q.empty()) {
		a = q.front();
		q.pop();
		for (int i = 0; i < 8; i++) {
			tx = a.x + dx[i];
			ty = a.y + dy[i];
			if (tx < 0 || tx > n || ty < 0 || ty>m
				|| mp[tx][ty] == '*' || v[tx][ty] == 1)
				continue;
			if (mp[tx][ty] == 'H') {
				ans = a.val + 1;
				break;
			}
			v[tx][ty] = 1;
			te.x = tx;
			te.y = ty;
			te.val = a.val + 1;
			q.push(te);
		}
		if (ans>0)break;
	}
	printf("%d\n", ans);

	return 0;
}