BFS(判断状态) HDOJ 3533 Escape

题目传送门

题意：一个人从(0, 0)逃往(n, m)，地图上有朝某个方向开炮的炮台，问最少逃脱步数

分析：主要在状态是否OK，当t时刻走到(x，y)，炮台是否刚好打中，因为只能是整数，所以用整除判断。题意不清楚，有些坑点。

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

const int N = 1e2 + 5;
struct Point	{
	int dir, t, v;		//N 1 E 2 S 3 W 4
}p[N][N];
struct Node	{
	int x, y, step;
	Node() {}
	Node(int x, int y, int step) : x (x), y (y), step (step) {}
};
int dx[5] = {-1, 1, 0, 0, 0};
int dy[5] = {0, 0, -1, 1, 0};
bool vis[N][N][N*10];
int n, m, k, d;

bool check(int x, int y, int tim)	{
	if (x < 0 || x > n || y < 0 || y > m || vis[x][y][tim] || p[x][y].dir != 0)	return false;
	else	return true;
}

bool check2(int x, int y, int tim)	{
	for (int i=x-1; i>=0; --i)	{		//up
		if (p[i][y].v == 0)	continue;
		if (p[i][y].dir != 3)	break;
		int dis = x - i;
		if (dis % p[i][y].v != 0)	break;
		int t = tim - dis / p[i][y].v;
		if (t % p[i][y].t == 0)	return false;
		else	break;
	}
	for (int i=x+1; i<=n; ++i)	{		//down
		if (p[i][y].v == 0)	continue;
		if (p[i][y].dir != 1)	break;
		int dis = i - x;
		if (dis % p[i][y].v != 0)	break;
		int t = tim - dis / p[i][y].v;
		if (t % p[i][y].t == 0)	return false;
		else	break;
	}
	for (int i=y-1; i>=0; --i)	{		//left
		if (p[x][i].v == 0)	continue;
		if (p[x][i].dir != 2)	break;
		int dis = y - i;
		if (dis % p[x][i].v != 0)	break;
		int t = tim - dis / p[x][i].v;
		if (t % p[x][i].t == 0)	return false;
		else	break;
	}
	for (int i=y+1; i<=m; ++i)	{		//right
		if (p[x][i].v == 0)	continue;
		if (p[x][i].dir != 4)	break;
		int dis = i - y;
		if (dis % p[x][i].v != 0)	break;
		int t = tim - dis / p[x][i].v;
		if (t % p[x][i].t == 0)	return false;
		else	break;
	}

	return true;
}

int BFS(void)	{
	Node s;
	s.x = s.y = s.step = 0;
	queue<Node> que;	que.push (s);
	vis[s.x][s.y][0] = true;
	while (!que.empty ())	{
		Node u = que.front ();	que.pop ();
		if (u.step > d)	continue;
		if (u.x == n && u.y == m && u.step <= d)	{
			return u.step;
		}
		for (int i=0; i<5; ++i)	{
			int tx = u.x + dx[i], ty = u.y + dy[i], tsp = u.step + 1;
			if (!check (tx, ty, tsp))	continue;
			if (!check2 (tx, ty, tsp))	continue;
			vis[tx][ty][tsp] = true;
			que.push (Node (tx, ty, tsp));
		}
	}

	return -1;
}

int main(void)	{
	map<char, int> mp;
	mp[‘N‘] = 1;	mp[‘E‘] = 2;	mp[‘S‘] = 3;	mp[‘W‘] = 4;
	while (scanf ("%d%d%d%d", &n, &m, &k, &d) == 4)	{
		memset (vis, false, sizeof (vis));
		memset (p, 0, sizeof (p));
		char str[3];	int t, v, x, y;
		getchar ();
		for (int i=0; i<k; ++i)	{
			scanf ("%s%d%d%d%d", &str, &t, &v, &x, &y);
			p[x][y].dir = mp[str[0]];
			p[x][y].t = t;	p[x][y].v = v;
		}
		int ans = BFS ();
		if (ans == -1)	puts ("Bad luck!");
		else printf ("%d\n", ans);
	}

	return 0;
}