Nightmare（HDU 1072）

题目大意：一般的搜索问题，用BFS能够0ms解决。有剩余时限让你去走迷宫，有一个特殊的东西可以让你的剩余时间重置。

具体的要求是：

1. We can assume the labyrinth is a 2 array.(二维迷宫)
2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too.（不能走墙）
3. If Ignatius get to the exit when the exploding time turns to 0, he can't get out of the labyrinth.（刚好在剩余0S走出也不行）
4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can't use the equipment to reset the bomb.（刚好在0s重置时间也不行）
5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish.（你可以使用重置若干次，不过在我的代码当中，是没考虑，不知道是不是测试数据太弱了，就那么AC了。）
6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.（一走到该道具，剩余时间立刻重置，不需要考虑时间）

测试数据：

3 3 3 2 1 1 1 1 0 1 1 3 4 8 2 1 1 0 1 1 1 0 1 0 4 1 1 0 4 1 1 0 0 0 0 0 0 1 1 1 1 4 1 1 1 3 5 8 1 2 1 1 1 1 1 4 1 0 0 0 1 0 0 1 1 4 1 0 1 1 0 1 1 0 0 0 0 3 0 1 1 1 4 1 1 1 1 1

4 -1 13

以下是我的AC代码：一般的BFS模板根据条件改编

#include
#include 
#include 
using namespace std;
short map[10][10], vis[10][10];
int Div[4][2] = { 1, 0, -1, 0, 0, 1, 0, -1 };
int n, m;
int sx, sy, ex, ey;
int times = -1;
struct pos
{
	int x;
	int y;
	int times;
	int Time;
	friend bool operator <(pos a, pos b)//使用优先队列，使得剩余时间最多的先出队
	{
		return a.Timeq;
	pos p, mid;
	p.x = sx;
	p.y = sy;
	p.times = 0;
	p.Time = 6;
	q.push(p);
	vis[sx][sy] = 1;
	while (!q.empty())
	{
		p = q.top();
		q.pop();
		if (p.x == ex&&p.y == ey&&p.Time > 0)
		{
			times = p.times;
			return;
		}
		if (p.Time == 1)//这个是针对0秒不可行，即当剩余时间为1S是还没走出，那永远不可以。
			continue;
		for (int i= 0; i < 4; i++)//这边加入一个判断，是为了判断该点附近有没有时间重置的地点。
		{
			mid = p;//具体理解看第三个测试数据吧，还好测试数据给的良心，不然完全不会想到。
			mid.x = p.x + Div[i][0];//道行还是太浅了。
			mid.y= p.y + Div[i][1];
			if (mid.x >= 1 && mid.x <= n&&mid.y <= m&&mid.y >= 1 && map[mid.x][mid.y] == 4&&p.Time<5)
			{
			mid.Time = 6;
			mid.times++;//本来想解释一下，但发现不会解释，看不懂的还是看测试数据三理解吧
			q.push(mid);
			p.Time = 5;
			p.times += 2;
				break;
			}
		}
		for (int i = 0; i < 4; i++)
		{
			mid = p;
			mid.x = p.x + Div[i][0];
			mid.y = p.y + Div[i][1];
			mid.times++;
			mid.Time--;
			if (mid.x>n || mid.x<1 || mid.y>m || mid.y < 1||!map[mid.x][mid.y]||vis[mid.x][mid.y])
				continue;
			if (map[mid.x][mid.y] == 4)
			{
				vis[mid.x][mid.y] = 1;
				mid.Time = 6;
				q.push(mid);
				continue;
			}
			else
			{
				vis[mid.x][mid.y] = 1;
				q.push(mid);
			}
		}
	}
}
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		cin >> n >> m;
		times = -1;
		for (int i = 1; i <= n;i++)
		for (int j = 1; j <= m; j++)
		{
			cin >> map[i][j];
			if (map[i][j] == 2)
			{
				sx = i; sy = j;
			}
			if (map[i][j] == 3)
			{
				ex = i; ey = j;
			}
		}
		bfs();
		cout << times << endl;
	}
}