BFS HDOJ 2102 A计划

题目传送门

题意:中文题面

分析:双层BFS,之前写过类似的题.总结坑点:

　　1.步数小于等于T都是YES　　2. 传送门的另一侧还是传送门或者墙都会死　　3. 走到传送门也需要一步

#include <bits/stdc++.h>
using namespace std;

char maze[2][11][11];
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int n, m, tot;
bool vis[2][11][11];
struct Point	{
	int x, y, z, step;
	Point ()	{}
	Point (int x, int y, int z, int step) : x (x), y (y), z (z), step (step) {}
};
Point s, e;

bool check(int x, int y, int z)	{
	if (x < 1 || x > n || y < 1 || y > m || vis[z][x][y] || maze[z][x][y] == '*')	return false;
	else	return true;
}

bool BFS(void)	{
	memset (vis, false, sizeof (vis));
	int res = 0x3f3f3f3f;
	queue<Point> que;	que.push (s);
	vis[s.z][s.x][s.y] = true;
	while (!que.empty ())	{
		Point u = que.front ();	que.pop ();
		if (u.x == e.x && u.y == e.y && u.z == e.z)	{
			res = min (res, u.step);	continue;
		}
		for (int i=0; i<4; ++i)	{
			int tx = u.x + dx[i], ty = u.y + dy[i], tz = u.z;
			if (!check (tx, ty, tz))	continue;
			if (maze[tz][tx][ty] == '#')	{
				if (maze[1-tz][tx][ty] == '*' || maze[1-tz][tx][ty] == '#' || vis[1-tz][tx][ty])	continue;
				vis[1-tz][tx][ty] = true;
				que.push (Point (tx, ty, 1 - tz, u.step + 1));
				continue;
			}
			vis[tz][tx][ty] = true;
			que.push (Point (tx, ty, tz, u.step + 1));
		}
	}
	return res <= tot;
}

int main(void)	{
	int T;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d%d%d", &n, &m, &tot);
		for (int k=0; k<2; ++k)	{
			if (k == 1)	getchar ();
			for (int i=1; i<=n; ++i)	{
				scanf ("%s", maze[k][i] + 1);
				for (int j=1; j<=m; ++j)	{
					if (maze[k][i][j] == 'S')	{
						s = Point (i, j, k, 0);
					}
					else if (maze[k][i][j] == 'P')	{
						e = Point (i, j, k, 0);
					}
				}
			}
		}
		if (BFS ())	puts ("YES");
		else	puts ("NO");
	}

	return 0;
}