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最新推荐文章于 2022-04-03 08:15:04 发布

原创最新推荐文章于 2022-04-03 08:15:04 发布 · 303 阅读

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本文介绍了一个使用广度优先搜索(BFS)算法解决迷宫最短路径问题的C++实现。通过创建一个二维数组来表示迷宫地图，并定义起点与终点，在确保地图中障碍物不可通过的情况下，实现了从起点到终点的最短步数计算。

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我的代码炸了。就是放着看看

#include <queue>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;
int N, M;
int Lu[1005][1005], Hui[1005][1005];
typedef pair<int, int> P;
int sx, sy, gx, gy;
const int INF = 0x3f3f3f3f;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int bfs()
{
	queue<P> que;
	for (int i = 0; i <= N + 1; i++)
		for (int j = 0; i <= M + 1; j++)
			Hui[i][j] = INF;
	que.push(P(sx, sy));
	Hui[sx][sy] = 0;
	while (que.size()){
		P p = que.front(); que.pop();
		if (p.first == gx && p.second == gy)
			break;
		for (int i = 0; i < 4; i++){
			int nx = p.first + dx[i], ny = p.second + dy[i];
			if (nx >= 0 && nx <= N + 1 && ny >= 0 && ny <= M + 1 && Lu[nx][ny] == 0 && Hui[nx][ny] == INF){
				que.push(P(nx, ny));
				Hui[nx][ny] = Hui[p.first][p.second] + 1;
			}
		}
	}
	return Hui[gx][gy];
}
int main()
{
	int n;
	int x1, x2, y1, y2;
	while (scanf("%d%d%*c", &N, &M) && (N,M)){
		for (int i = 1; i <= N; i++)
			for (int j = 1; j <= M; j++)
				scanf("%d", &Lu[i][j]);
		scanf("%d", &n);
		while (n--){
			scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
			if (Lu[x1][y1] == 0 || Lu[x2][y2] == 0){
				printf("NO\n");
				continue;
			}
			int Min = abs(x1 - x2) + abs(y1 - y2);
			for (int i = x1 - 1; i <= x2 + 1; i++){
				Lu[i][y1 -1] = 0;
				Lu[i][y2 + 1] = 0;
			}
			for (int j = y1 - 1; j <= y2 + 1; j++){
				Lu[x1 - 1][j] = 0;
				Lu[x2 + 1][j] = 0;
			}
			sx = x1;sy = y1;gx = x2;gy = y2;
			bfs();
			if (Min == Hui[gx][gy])
				printf("YES\n");
			else
				printf("NO\n");
		}
	}
	return 0;
}