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最新推荐文章于 2021-02-24 14:48:07 发布

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本文介绍了一个基于广度优先搜索算法的游戏场景应用，通过定义骑士的移动规则和地图上的障碍物，实现从起点到终点的路径寻找。使用C++语言实现了算法，并详细展示了如何遍历地图，避免重复访问和无效路径，最终找到从骑士位置到宝藏位置的最短步数。

#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;
}