BFS(双向) HDOJ 3085 Nightmare Ⅱ

题目传送门

题意:一个人去救女朋友,两个人都在运动,还有鬼在"扩散",问最少几秒救到女朋友

分析:开两个队列来表示两个人走过的路,一个人走到的地方另一个人已经vis了,那么就是相遇了,鬼就用曼哈顿距离判断.

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

const int N = 8e2 + 5;
char maze[N][N];
int n, m;
bool vis[N][N][2];
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
struct Point	{
	int x, y;
	Point()	{}
	Point(int x, int y)	: x (x), y (y) {}
};
queue<Point> que[2];
Point mm, gg, gho[2];

int get_dis(int x, int y, int ex, int ey)	{
	return abs (ex - x) + abs (ey - y);
}

bool check2(int x, int y, int tim)	{
	for (int i=0; i<2; ++i)	{
		if (get_dis (x, y, gho[i].x, gho[i].y) <= 2 * tim)	return false;
	}
	return true;
}

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

void init(void)	{
	while (!que[0].empty ())	que[0].pop ();
	while (!que[1].empty ())	que[1].pop ();
	memset (vis, false, sizeof (vis));
}

bool BFS(int typ, int tim)	{
	int sz = que[typ].size ();
	while (sz--)	{
		Point u = que[typ].front ();	que[typ].pop ();
		if (!check (u.x, u.y, typ) || !check2 (u.x, u.y, tim))	continue;
		for (int i=0; i<4; ++i)	{
			int tx = u.x + dx[i], ty = u.y + dy[i];
			if (!check (tx, ty, typ) || !check2 (tx, ty, tim))	continue;
			if (vis[tx][ty][typ^1])	return true;
			if (vis[tx][ty][typ])	continue;
			vis[tx][ty][typ] = true;
			que[typ].push (Point (tx, ty));
		}
	}
	return false;
}

int run(void)	{
	init ();
	vis[mm.x][mm.y][0] = true;
	vis[gg.x][gg.y][1] = true;
	que[0].push (Point (mm.x, mm.y));
	que[1].push (Point (gg.x, gg.y));
	int step = 0;
	while (!que[0].empty () || !que[1].empty ())	{
		++step;
		for (int i=0; i<3; ++i)	{
			if (BFS (0, step))	return step;
		}
		if (BFS (1, step))	return step;
	}
	return -1;
}

int main(void)	{
	int T;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d%d", &n, &m);
		for (int i=1; i<=n; ++i)	{
			scanf ("%s", maze[i] + 1);
		}
		int t = 0;
		for (int i=1; i<=n; ++i)	{
			for (int j=1; j<=m; ++j)	{
				if (maze[i][j] == 'M')	{
					mm.x = i;	mm.y = j;
				}
				else if (maze[i][j] == 'G')	{
					gg.x = i;	gg.y = j;
				}
				else if (maze[i][j] == 'Z')	{
					gho[t].x = i;	gho[t++].y = j;
				}
			}
		}
		printf ("%d\n", run ());
	}

	return 0;
}