Description

Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not be filled with rock. You can move north, south, east or west one cell at a step. These moves are called walks.
One of the empty cells contains a box which can be moved to an adjacent free cell by standing next to the box and then moving in the direction of the box. Such a move is called a push. The box cannot be moved in any other way than by pushing, which means that if you push it into a corner you can never get it out of the corner again.One of the empty cells is marked as the target cell. Your job is to bring the box to the target cell by a sequence of walks and pushes. As the box is very heavy, you would like to minimize the number of pushes. Can you write a program that will work out the best such sequence?

Input

Output

Sample Input

1 7
SB....T
1 7
SB..#.T
7 11
###########
#T##......#
#.#.#..####
#....B....#
#.######..#
#.....S...#
###########
8 4
....
.##.
.#..
.#..
.#.B
.##S
....
###T
0 0

Sample Output

Maze #1
EEEEE

Maze #2
Impossible.

Maze #3
eennwwWWWWeeeeeesswwwwwwwnNN

Maze #4
swwwnnnnnneeesssSSS

Source

Solution:

一道非常不错的 A* 大水题。

采用 Manhattan 距离作为估价函数，将 push 操作和 walk 操作分开评估：

• push: h() = Manhattan(box, destination)。
• walk: h() = Manhattan(man, box) - 1，减 1 的原因是 man 只要走到 box 四周的某一格即可。

此题的“关闭列表”改为刚得到状态时直接记录最小代价，而不是当状态从开启列表中移除后记录，可以减少入队次数从而提升时空效率。

路径记录数组一定要开得足够大。

Code: O(玄学) [2324K, 79MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int dx[4] = {0, 1, 0,-1};
const int dy[4] = {1, 0,-1, 0};
const char DIR[4] = {'E', 'S', 'W', 'N'};

int cas = 0, R, C;
char mp[22][22];

struct Point{
	int x, y;
	
	inline bool operator == (const Point &p2) const {return x == p2.x && y == p2.y;}
	
	inline bool isLegal() {return x > 0 && x <= R && y > 0 && y <= C && mp[x][y] != '#';}
} T;

struct State{
	Point man, box;
	int pf, pg, ph, wf, wg, wh;
	int id;
	
	inline bool operator < (const State &st2) const {
		if(pf != st2.pf) return pf > st2.pf;
		return wf > st2.wf;
	}
} S;

priority_queue<State> openList;
int closeList[22][22][22][22][2];

inline void clearPQ(priority_queue<State> &pq) {priority_queue<State> tmp; swap(tmp, pq);}

inline int Manhattan(const Point &p1, const Point &p2) {return abs(p1.x - p2.x) + abs(p1.y - p2.y);}

inline bool isBetter(const State &st){
	int *buf = closeList[st.man.x][st.man.y][st.box.x][st.box.y];
	if(buf[0] != st.pf) return st.pf < buf[0];
	return st.wf < buf[1];
}

inline void updateClose(const State &st){
	int *buf = closeList[st.man.x][st.man.y][st.box.x][st.box.y];
	buf[0] = st.pf, buf[1] = st.wf;
}

int fromId[100001], topfrom;
char fromDir[100001];

inline int Astar(){
	S.pg = 0, S.ph = Manhattan(S.box, T), S.pf = S.pg + S.ph;
	S.wg = 0, S.wh = Manhattan(S.man, S.box) - 1, S.wf = S.wg + S.wh;
	S.id = 0;
	openList.push(S);
	while(!openList.empty()){
		State u = openList.top(), v;
		openList.pop();
		for(register int i = 0; i < 4; i++){
			v = u, v.man.x += dx[i], v.man.y += dy[i];
			if(!v.man.isLegal()) continue;
			if(v.man == v.box){  // Push
				v.box.x += dx[i], v.box.y += dy[i];
				if(!v.box.isLegal()) continue;
				v.pg++, v.ph = Manhattan(v.box, T), v.pf = v.pg + v.ph;
				if(!isBetter(v)) continue;
				fromId[v.id = ++topfrom] = u.id, fromDir[v.id] = DIR[i];
				openList.push(v);
				updateClose(v);
				if(v.box == T) return v.id;
			}
			else{  // Walk
				v.wg++, v.wh = Manhattan(v.man, v.box) - 1, v.wf = v.wg + v.wh;
				if(!isBetter(v)) continue;
				fromId[v.id = ++topfrom] = u.id, fromDir[v.id] = DIR[i] - 'A' + 'a';
				openList.push(v);
				updateClose(v);
			}
		}
	}
	return -1;
}

inline void traceBack(int cur){
	if(fromId[cur]) traceBack(fromId[cur]);
	putchar(fromDir[cur]);
}

int main(){
	while(scanf("%d%d", &R, &C) != EOF && R && C){
		for(register int i = 1; i <= R; i++){
			scanf("%s", mp[i] + 1);
			for(register int j = 1; j <= C; j++){
				if(mp[i][j] == 'T') T.x = i, T.y = j;
				if(mp[i][j] == 'S') S.man.x = i, S.man.y = j;
				if(mp[i][j] == 'B') S.box.x = i, S.box.y = j;
			}
		}
		clearPQ(openList), memset(closeList, 0x3f, sizeof(closeList));
		topfrom = 0;
		int endId = Astar();
		printf("Maze #%d\n", ++cas);
		if(endId == -1) puts("Impossible.\n");
		else traceBack(endId), puts("\n");
	}
	return 0;
}