sicily 1215 脱离地牢 bfs

        本认为自己掌握了bfs，今天下午被这题卡了，以前做的迷宫都是单独一个人为状态的迷宫，队列保存的状态就是一个坐标，而这题就两个人，两个状态量，一开始我以一个人为状态量，WA好几次，后来发现自己之前对bfs有些误解， bfs的队列是保存每一次结果的状态，谨记谨记！！

#include <iostream>
#include <queue>
#include <memory.h>
using namespace std;

int n, m;

//状态
struct state
{
	int P_x, P_y, H_x, H_y;
	int step;
	bool ismeet() {
		return P_x == H_x && P_y == H_y;
	}//判断两个是否到达同一个
	bool isvalid() {
		return P_x > 0 && P_x <= n &&  P_y > 0 && P_y <= m
		&& H_x > 0 && H_x <= n && H_y > 0 && H_y <= m ;
	}//判断坐标是否合法
	bool iscollision(state s)
	{
		return P_x == s.H_x && P_y == s.H_y && H_x == s.P_x && H_y == s.P_y;
	}//检测两个人是否相遇
	state(int P_x, int P_y, int H_x, int H_y) {
		this->H_x = H_x; this->H_y = H_y; this->P_x = P_x; this->P_y = P_y; step = 0;
	}
	state() {
		step = 0;
	}
};

enum direction{N, S, W, E};

//N,S,W,E移动方向
int mov[4][2] = {{-1, 0},{1, 0}, {0, -1}, {0, 1}};
int match[4];
char maze[21][21];
bool isvisit[21][21][21][21];

int P_x, P_y, H_x, H_y;

void bfs();

int main()
{
	char c;
	while (cin >> n >> m)
	{
		memset(isvisit, false, sizeof(isvisit));

		for (int i = 1; i <= n; i++)	
			for (int j = 1; j <= m; j++)
			{
				cin >> maze[i][j];
				if (maze[i][j] == 'P')
				{
					P_x = i;
					P_y = j;
					maze[i][j] = '.';
				}
				if (maze[i][j] == 'H')
				{
					H_x = i;
					H_y = j;
					maze[i][j] = '.';
				}
			}

			//匹配方向
			for (int i = 0; i < 4; i++)
			{
				cin >> c;
				switch (c)
				{
				case 'N':
					match[i] = N;
					break;
				case 'S':
					match[i] = S;
					break;
				case 'W':
					match[i] = W;
					break;
				case 'E':
					match[i] = E;
					break;
				}
			}
			bfs();
	}
	return 0;
}

void bfs()
{
	queue <state> Q;
	Q.push(state(P_x, P_y, H_x, H_y));

	state tmp, hold;

	while (!Q.empty())
	{
		tmp = Q.front();
		Q.pop();
		if (tmp.step > 255)
		{
			cout << "Impossible" << endl;
			return;
		}
		//从N,S,W,E开始更新状态
		for (int i = 0; i < 4; i++)
		{
			hold.P_x = tmp.P_x + mov[i][0];
			hold.P_y = tmp.P_y + mov[i][1];
			hold.H_x = tmp.H_x + mov[match[i]][0];
			hold.H_y = tmp.H_y + mov[match[i]][1];
			hold.step = tmp.step + 1;
		
			if (hold.isvalid() && 
				maze[hold.H_x][hold.H_y] != '!' && 
				maze[hold.P_x][hold.P_y] == '.' ) //每次把合法的状态放到队列
			{
				if (maze[hold.H_x][hold.H_y] == '#')
				{
					hold.H_x -= mov[match[i]][0];
					hold.H_y -= mov[match[i]][1];
				}
				if (!isvisit[hold.P_x][hold.P_y][hold.H_x][hold.H_y])
				{
					isvisit[hold.P_x][hold.P_y][hold.H_x][hold.H_y] = true;
					if (hold.ismeet() || tmp.iscollision(hold))
					{
						cout << hold.step << endl;
						return ;
					}
					else
						Q.push(hold);
				}
			}
		}

	}
	cout << "Impossible" << endl;
}