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;

}