每个测试输入包含1个测试用例 第一行输入两个数字N,M表示地图的大小。其中0<N,M<=8。 接下来有N行,每行包含M个字符表示该行地图。其中 . 表示空地、X表示玩家、*表示箱子、#表示障碍、@表示目的地。 每个地图必定包含1个玩家、1个箱子、1个目的地。
输出一个数字表示玩家最少需要移动多少步才能将游戏目标达成。当无论如何达成不了的时候,输出-1。
4 4 .... ..*@ .... .X.. 6 6 ...#.. ...... #*##.. ..##.# ..X... .@#...
3 11
import sys import collections def moveBox(bmap, N, M): dic = ((1,0), (-1,0), (0,1), (0,-1)) for i in range(N): for j in range(M): if bmap[i][j] == '*': box = [i, j] if bmap[i][j] == 'X': people = [i, j] dp = [[[[False for _ in range(M)] for _ in range(N)] for _ in range(M)] for _ in range(N)] dp[box[0]][box[1]][people[0]][people[1]] = True stack = collections.deque([[box[0], box[1], people[0], people[1]]]) step = 0 while stack: step += 1 lenth = len(stack) for _ in range(lenth): bx, by, px, py = stack.popleft() for dx, dy in dic: nx, ny = px + dx, py + dy # can move if -1 < nx < N and -1 < ny < M and bmap[nx][ny] != '#': # touch the box, move the box if nx == bx and ny == by: nbx, nby = bx+dx, by+dy # the box is in range if -1 < nbx < N and -1 < nby < M: # arrive the target if bmap[nbx][nby] == '@': return step else: # first arrive, keep the state if not dp[nbx][nby][nx][ny]: dp[bx+dx][by+dy][nx][ny] = True stack.append([bx+dx, by+dy, nx, ny]) else: # keep the state if not dp[bx][by][nx][ny]: dp[bx][by][nx][ny] = True stack.append([bx, by, nx, ny]) return -1 if __name__ == '__main__': sin = sys.stdin #sin = open('move_box.txt') while True: line = sin.readline() if not line: break ## line = line.splitlines()[0] N, M = [int(t) for t in line.split()] bmap = [] for _ in range(N): bmap.append([ch for ch in sin.readline().splitlines()[0]]) print(moveBox(bmap, N, M))
开个四维vis[x][y][xb][yb]代表人在(x,y),箱子在(xb,yb)这个状态的最小步数。 #include<iostream> #include<stdio.h> #include<queue> using namespace std; struct q{int x,y,xb,yb;q(int x,int y,int xb,int yb):x(x),y(y),xb(xb),yb(yb){}};int a[]={0,0,1,-1},b[]={1,-1,0,0};char mp[10][10];int vis[10][10][10][10];int bx,by,sx,sy,ex,ey,n,m;int bfs(){vis[sx][sy][bx][by]=1;q p(sx,sy,bx,by);queue<q> que;que.push(p);while(que.size()){p=que.front();que.pop();if(p.xb==ex&&p.yb==ey)return vis[p.x][p.y][p.xb][p.yb]-1;for(int i=0;i<4;i++){int nx=p.x+a[i],ny=p.y+b[i];if(nx<0||ny<0||mp[nx][ny]=='#'||nx>=n||ny>=m)continue;if(nx==p.xb&&ny==p.yb){if(nx+a[i]<0||ny+b[i]<0||mp[nx+a[i]][ny+b[i]]=='#'||nx+a[i]>=n||ny+b[i]>=m)continue;if(vis[nx][ny][nx+a[i]][ny+b[i]])continue;vis[nx][ny][nx+a[i]][ny+b[i]]=vis[p.x][p.y][p.xb][p.yb]+1;que.push(q(nx,ny,nx+a[i],ny+b[i]));}else{if(vis[nx][ny][p.xb][p.yb])continue;vis[nx][ny][p.xb][p.yb]=vis[p.x][p.y][p.xb][p.yb]+1;que.push(q(nx,ny,p.xb,p.yb));}}}return -1;}int main(){scanf("%d%d",&n,&m);for(int i=0;i<n;i++)scanf("%s",mp[i]);for(int i=0;i<n;i++)for(int j=0;j<m;j++)if(mp[i][j]=='*')bx=i,by=j;else if(mp[i][j]=='X')sx=i,sy=j;else if(mp[i][j]=='@')ex=i,ey=j;cout<<bfs()<<endl;return 0;}
做法是,只有人是主动的有四个方向。
情况1:如果人走到了一个点,而且这个点恰好是箱子,那么箱子自然也要与人同样的方向移动。 此时需要判断人和箱子的位置是否都满足 既不是障碍,也不出图
情况2:只有人走了一个点,这个点不是箱子,那么自然不需要判断箱子的情况了。
开个四维数组, 两维给人记录X,Y, 另外两维给箱子记录X,Y。
这样就是裸的bfs
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
char mp[10][10];
int vis[10][10][10][10];
int data[4][2]={0,1,0,-1,1,0,-1,0};
int n,m;
struct node{
int x,y,step;
int xx,yy;
node(int _x,int _y,int _step,int _xx,int _yy){x=_x,y=_y,step=_step,xx=_xx,yy=_yy;}
};
int bfs(int sxx,int syy,int sx,int sy){
queue<node> q;
q.push(node(sx,sy,0,sxx,syy));
while(!q.empty())
{
node s=q.front();
q.pop();
vis[s.x][s.y][s.xx][s.yy]=1;
//cout<<s.x <<" "<<s.y<<" "<<s.xx<<" "<<s.yy<<" "<<s.step<<endl;
if(mp[s.xx][s.yy]=='@') return s.step;
for(int i=0;i<4;i++){
node tmp=s;
tmp.step++;
tmp.x+=data[i][0];
tmp.y+=data[i][1];
if(tmp.x>n||tmp.y>m||tmp.x<1||tmp.y<1) continue;
if(mp[tmp.x][tmp.y]=='#') continue;
if(tmp.x==tmp.xx&&tmp.y==tmp.yy)
tmp.xx+=data[i][0],tmp.yy+=data[i][1];
if(tmp.xx>n||tmp.yy>m||tmp.xx<1||tmp.yy<1) continue;
if(mp[tmp.xx][tmp.yy]=='#') continue;
if(vis[tmp.x][tmp.y][tmp.xx][tmp.yy]!=1)
q.push(tmp);
}
}
return -1;
}
int main(){
cin>>n>>m;
for(int i=1;i<=n;i++)
cin>>&mp[i][1];
int sx,sy,xx,yy;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(mp[i][j]=='X')
sx=i,sy=j;
if(mp[i][j]=='*')
xx=i,yy=j;
}
}
cout<<bfs(xx,yy,sx,sy)<<endl;
}
/*
3 3
#.X
#*.
@..
*/
参考@周昆的代码,加了一些注释便于大家理解
# 前两位表示箱子的起始位置,后两位表示玩家的位置start = [-1] * 4 end = [-1] * 2 n, m = map(int, input().split()) mat = list() for i in range(n): line = input() mat.append(line) for j in range(len(line)): # box pos if line[j] == '*': start[0], start[1] = i, j # player pos if line[j] == 'X': start[2], start[3] = i, j # dest pos if line[j] == '@': end[0], end[1] = i, j # 四维表用以记录箱子的x,y及玩家的x,y reach = [[[[-1 for _ in range(m)] for _ in range(n)] for _ in range(m)] for _ in range(n)] queue = list() queue.append(start) direction = ((0, 1), (1, 0), (0, -1), (-1, 0)) reach[start[0]][start[1]][start[2]][start[3]] = 0 while len(queue): cur = queue.pop(0) if cur[0] == end[0] and cur[1] == end[1]: print(reach[cur[0]][cur[1]][cur[2]][cur[3]]) break # 可以往四个方向推 for i in range(4): player_x = cur[2] + direction[i][0] player_y = cur[3] + direction[i][1] # 检查箱子和玩家位置是否合法 if 0 <= player_x < n and 0 <= player_y < m and mat[player_x][player_y] != '#': # 检查玩家和箱子是否重合 if player_x == cur[0] and player_y == cur[1]: # 重合的时候将箱子往同一个方向推并检查箱子状态是否合法 box_x, box_y = cur[0] + direction[i][0], cur[1] + direction[i][1] if box_x < 0 or box_x >= n or box_y < 0 or box_y >= m or mat[box_x][box_y] == '#': continue else: # 如果不重合则箱子位置不动 box_x, box_y = cur[0], cur[1] # 判断该点是否已经遍历过 if reach[box_x][box_y][player_x][player_y] < 0: queue.append([box_x, box_y, player_x, player_y]) reach[box_x][box_y][player_x][player_y] = reach[cur[0]][cur[1]][cur[2]][cur[3]] + 1 if cur[0] != end[0] or cur[1] != end[1]: print(-1)
from collections import deque start = [0] * 4 end = [0] * 2 N, M = map(int, raw_input().split()) maze = [] for i in range(N): maze.append([]) for j, s in enumerate(raw_input()): maze[i].append(s) if s == '*': start[0], start[1] = i, j elif s == 'X': start[2], start[3] = i, j elif s =='@': end[0], end[1] = i, j # box_x, box_y, per_x, per_y reach = [[[[-1 for _ in range(M)] for _ in range(N)] for _ in range(M)] for _ in range(N)] mazeQueue = deque() mazeQueue.append(start) direction = ((0, 1), (1, 0), (0, -1), (-1, 0)) reach[start[0]][start[1]][start[2]][start[3]] = 0 while len(mazeQueue): point = mazeQueue.popleft() if point[0] == end[0] and point[1] == end[1]: print reach[point[0]][point[1]][point[2]][point[3]] break for i in range(4): # new coordinate for person xper = point[2]+direction[i][0] yper = point[3]+direction[i][1] # check validity if 0 <= xper < N and 0 <= yper < M and maze[xper][yper] != "#": if xper == point[0] and yper == point[1]: xbox, ybox = point[0] + direction[i][0], point[1] + direction[i][1] if xbox < 0 or xbox >= N or ybox < 0 or ybox >= M or maze[xbox][ybox] == '#': continue else: xbox, ybox = point[0], point[1] if reach[xbox][ybox][xper][yper] < 0: mazeQueue.append([xbox, ybox, xper, yper]) reach[xbox][ybox][xper][yper] = reach[point[0]][point[1]][point[2]][point[3]] + 1 if point[0] != end[0] or point[1] != end[1]: print -1
import java.util.ArrayList; import java.util.LinkedList; import java.util.Queue; import java.util.Scanner; /** * Created by Administrator on 2017/3/16. */ class Stage{ public Stage(int peoplex, int peopley) { this.peoplex = peoplex; this.peopley = peopley; } public Stage(int peoplex, int peopley, int boxx, int boxy) { this.peoplex = peoplex; this.peopley = peopley; this.boxx = boxx; this.boxy = boxy; } @Override public String toString() { return "people:("+this.peoplex+","+this.peopley+") box:("+this.boxx+","+this.boxy+")"; } public Stage() { } int peoplex; int peopley; int boxx; int boxy; Stage previous; public boolean equals(Object obj) { if (obj instanceof Stage) { if (((Stage) obj).peoplex==peoplex&&((Stage) obj).peopley==peopley&&((Stage) obj).boxx==boxx&&((Stage) obj).boxy==boxy)return true; } return false; } } public class test12 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while (sc.hasNext()){ int n = sc.nextInt(); int m = sc.nextInt(); char[][] map = new char[n][m]; for (int i=0;i<n;i++){ map[i] = sc.next().toCharArray(); } int peoplex=0; int peopley=0; int boxx=0; int boxy=0; for (int i=0;i<n;i++){ for (int j=0;j<m;j++){ if (map[i][j]=='X'){ peoplex=i; peopley=j; } if (map[i][j]=='*'){ boxx=i; boxy=j; } } } Stage begin = new Stage(peoplex,peopley,boxx,boxy); System.out.println(begin); Stage result = BFS(begin,map,m,n); if (result==null){ System.out.println(-1); }else { while (result.previous!=null){ System.out.println(result.previous); result = result.previous; } } } } public static Stage BFS(Stage begin,char[][] map,int m,int n){ Queue<Stage> queue = new LinkedList<Stage>(); ArrayList<Stage> arrayList = new ArrayList<Stage>(); int[][] dir = {{1,0},{0,1},{-1,0},{0,-1}}; queue.offer(begin); while(!queue.isEmpty()){ Stage local = queue.remove(); arrayList.add(local); for(int i=0; i< dir.length; i++){ Stage next = new Stage(local.peoplex+dir[i][0], local.peopley+dir[i][1]); next.previous = local; if (next.peoplex>=0&&next.peoplex<m&&next.peopley<n&&next.peopley>=0&&map[next.peoplex][next.peopley]!='#'){ if (next.peoplex==local.boxx&&next.peopley==local.boxy){ next.boxx=local.boxx+dir[i][0]; next.boxy=local.boxy+dir[i][1]; }else { next.boxx=local.boxx; next.boxy=local.boxy; } if (arrayList.contains(next))continue; if (next.boxx>=0&&next.boxx<m&&next.boxy<n&&next.boxy>=0&&map[next.boxx][next.boxy]!='#'){ arrayList.add(next); queue.offer(next); }else { continue; } if (map[next.boxx][next.boxy]=='@'){ return next; } } } } return null; } }此答案利用BFS计算,最后输出最短的走法
import java.util.*; public class Main { private static class State { int px, py, bx, by; State pre; public State(int px, int py, int bx, int by, State pre) { this.px = px; this.py = py; this.bx = bx; this.by = by; this.pre = pre; } } public static void main(String[] args) { Scanner sc = new Scanner(System.in); while (sc.hasNext()) { String s = sc.nextLine(); int n = Integer.parseInt(s.split(" ")[0]); int px = -1, py = -1, bx = -1, by = -1; char[][] maze = new char[n][]; for (int i = 0; i < n; i++) { maze[i] = sc.nextLine().toCharArray(); for (int j = 0; j < maze[i].length; j++) { if (maze[i][j] == 'X') { px = i; py = j; } else if (maze[i][j] == '*') { bx = i; by = j; } } } State start = new State(px, py, bx, by, null); List<State> list = bfs(maze, start); System.out.println(list.size() - 1); } } private static List<State> bfs(char[][] maze, State start) { int n = maze.length; int m = maze[0].length; boolean[][][][] added = new boolean[n][m][n][m]; Queue<State> queue = new LinkedList<>(); LinkedList<State> list = new LinkedList<>(); queue.add(start); added[start.px][start.py][start.bx][start.py] = true; int[][] move = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; State end = null; while (!queue.isEmpty()) { State cur = queue.poll(); if (maze[cur.bx][cur.by] == '@') { end = cur; break; } for (int[] a : move) { State next = new State(cur.px + a[0], cur.py + a[1], cur.bx, cur.by, cur); if (next.px == next.bx && next.py == next.by) { next.bx += a[0]; next.by += a[1]; if (next.bx < 0 || next.bx >= n || next.by < 0 || next.by >= m || maze[next.bx][next.by] == '#') continue; } else if (next.px < 0 || next.px >= n || next.py < 0 || next.py >= m || maze[next.px][next.py] == '#') { continue; } if (!added[next.px][next.py][next.bx][next.by]) { queue.add(next); added[next.px][next.py][next.bx][next.by] = true; } } } if (end != null) { while (end != null) { list.addFirst(end); end = end.pre; } } return list; } }
人动,
以人作bfs起点
人在不越界不撞墙的情况下,有两种状态
撞箱子,不撞箱子,
撞箱子,
箱子在不越界不撞墙的情况下,
会以人的方向前进。
字符串输入到二维字符数组可以一行一行的输入
char map[10][10];
for ( int i = 1; i n; ++ i )
cin >> &map[i][1];
可以用四维数组存放人和箱子共同的状态以表示是否之前人和箱子到过此状态
int vis[10][10][10][10];
四个方向
纵轴的x加一或减一,左右平移为加0.所以为
int x[] = {0,0,1,-1};
横轴的y与x同理,但对应位置上不能同时存在1和-1,因为是上下左右,直上直下。,所以为
int y[] = {1,-1,0,0};
入队 则将其vis置为访问过,防止重复入队。
相较 出队列再检查是否到终点,入队列即检查应该比较快
因为队列可能很长,如果入队检查已经找到了,则不需要将之前的出队,节约了时间
出队检查相较入队检查只多检查了一个队头的节点。
#include <iostream> #include <queue> using namespace std; char map[10][10]; int vis[10][10][10][10]; int x[] = {0,0,1,-1}; int y[] = {1,-1,0,0}; int n, m; struct Point{ int _px, _py; int _bx, _by; int _step; Point(int px, int py, int step, int bx, int by ):_px(px),_py(py),_bx(bx),_by(by),_step(step) {} }; int bfs(int px, int py, int bx, int by) { if ( map[bx][by] == '@ ' ) return 0; queue <Point> Q; Q.push(Point(px,py,0,bx,by)); vis[px][py][bx][by] = 1; while ( !Q.empty() ) { Point point = Q.front(); Q.pop(); for ( int i = 0; i < 4; i ++ ) { Point next = point; ++next._step; next._px += x[i]; next._py += y[i]; //人是否越界 if ( next._px < 1 || next._px > n || next._py < 1 || next._py > m ) continue; //人是否撞墙 if ( map[next._px][next._py] == '#' ) continue; //人是否撞到箱子 if ( next._px == next._bx && next._py == next._by ) next._bx += x[i], next._by += y[i]; //箱子是否到终点 if ( map[next._bx][next._by] == '@ ' ) //检查是否到终点 return point._step; //箱子是否越界 if ( next._bx < 1 || next._bx > n || next._by < 1 || next._by > m ) continue; //箱子是否撞墙 if ( map[next._bx][next._by] == '#' ) continue; //是否访问过 if ( vis[next._px][next._py][next._bx][next._by] != 1 ) {//置标记 vis[next._px][next._py][next._bx][next._by] = 1; Q.push(next); } } } return -1; } int main() { cin >> n >> m; for ( int i = 1; i <= n; ++ i ) cin >> &map[i][1]; int px, py, bx, by; for ( int i = 1; i <= n; ++ i ) for ( int j = 1; j <= n; ++ j ) { if ( map[i][j] == 'X' ) { px = i; py = j; } else if ( map[i][j] == '*' ) { bx = i; by = j; } } cout << bfs(px,py,bx,by); return 0; }
import java.util.*; public class Main{ public static void main(String...args){ Scanner scan = new Scanner(System.in); int N = scan.nextInt(); int M = scan.nextInt(); char[][] map = new char[N+2][M+2]; boolean[][]used = new boolean[N+2][M+2]; for(int i=0;i<map.length;i++){ Arrays.fill(map[i], '#'); } int[] start = null; int[] end = null; int[] box = null; for(int i=1;i<map.length-1;i++){ int j=1; for(char ch: scan.next().toCharArray()) { map[i][j] = ch; if (ch == 'X') { start = new int[]{i, j}; } if (ch == '@') { end = new int[]{i, j}; } if (ch == '*') { box = new int[]{i, j}; } j++; } } int res = findWay(map, used, start, end, box, 0); System.out.println(res); } private static int findWay(char[][]map, boolean[][]used, int[] start, int[] end, int[] box, int wayLen){ if(box[0] == end[0] && box[1] == end[1]){ return wayLen; } else{ int res = Integer.MAX_VALUE; int[]curbox = box.clone(); int[][]dirs = new int[][]{{0, -1}, {-1, 0}, {0, 1}, {1, 0}};//left, up, right, down; for(int[] dir: dirs){ int[] next = new int[]{start[0]+dir[0], start[1]+dir[1]}; if(used[next[0]][next[1]] || map[next[0]][next[1]] == '#'){ continue; } if(next[0] == box[0] && next[1] == box[1]){ curbox = new int[]{box[0]+dir[0], box[1]+dir[1]}; if(used[curbox[0]][curbox[1]] || map[curbox[0]][curbox[1]] == '#'){ continue; } } used[next[0]][next[1]] = true; //System.out.println("next: "+Arrays.toString(next)); int curLen = findWay(map, used, next, end, curbox, wayLen+1); used[next[0]][next[1]] = false; res = curLen == -1? res: Math.min(res, curLen); } return res == Integer.MAX_VALUE? -1: res; } } }
#include <bits/stdc++.h> using namespace std; #define MP make_pair #define X first #define Y second using Coord = pair<int, int>; using St = pair<Coord, Coord>; using Maze = vector<vector<char>>; St findPlayerAndBox(Maze &maze, int N, int M) { Coord player, box; for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) { if (maze[i][j] == 'X') player = MP(i, j); if (maze[i][j] == '*') box = MP(i, j); } return MP(player, box); } Coord findBoxTarget(Maze &maze, int N, int M) { for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) if (maze[i][j] == '@') return MP(i, j); return MP(0, 0); } vector<St> allPossibleOutcomes(Maze &maze, St state, int N, int M) { St cur_st = state; Coord player = cur_st.X; Coord box = cur_st.Y; vector<St> result; // left if (player.Y - 1 == box.Y && player.X == box.X) { // box in the left if (box.Y >= 1 && maze[box.X][box.Y - 1] != '#') { Coord player_ = MP(player.X, player.Y - 1); Coord box_ = MP(box.X, box.Y - 1); result.push_back(MP(player_, box_)); } } else if (player.Y - 1 >= 0 && maze[player.X][player.Y - 1] != '#') { Coord player_ = MP(player.X, player.Y - 1); result.push_back(MP(player_, box)); } // right if (player.Y + 1 == box.Y && player.X == box.X) { // box in the right if (box.Y < M - 1 && maze[box.X][box.Y + 1] != '#') { Coord player_ = MP(player.X, player.Y + 1); Coord box_ = MP(box.X, box.Y + 1); result.push_back(MP(player_, box_)); } } else if (player.Y < M - 1 && maze[player.X][player.Y + 1] != '#') { Coord player_ = MP(player.X, player.Y + 1); result.push_back(MP(player_, box)); } // up if (player.X - 1 == box.X && player.Y == box.Y) { // box up if (box.X >= 1 && maze[box.X - 1][box.Y] != '#') { Coord player_ = MP(player.X - 1, player.Y); Coord box_ = MP(box.X - 1, box.Y); result.push_back(MP(player_, box_)); } } else if (player.X >= 1 && maze[player.X - 1][player.Y] != '#') { Coord player_ = MP(player.X - 1, player.Y); result.push_back(MP(player_, box)); } // down if (player.X + 1 == box.X && player.Y == box.Y) { // box below if (box.X < N - 1 && maze[box.X + 1][box.Y] != '#') { Coord player_ = MP(player.X + 1, player.Y); Coord box_ = MP(box.X + 1, box.Y); result.push_back(MP(player_, box_)); } } else if (player.X < N - 1 && maze[player.X + 1][player.Y] != '#') { Coord player_ = MP(player.X + 1, player.Y); result.push_back(MP(player_, box)); } return result; } int main(void) { int N, M; cin >> N >> M; Maze maze; for (int i = 0; i < N; i++) { vector<char> row; for (int j = 0; j < M; j++) { char ch; cin >> ch; row.push_back(ch); } maze.push_back(row); } // (player, box) queue<St> q; map<St, int> steps; set<St> vis; St initial = findPlayerAndBox(maze, N, M); vis.insert(initial); steps.insert(MP(initial, 0)); q.push(initial); Coord target_box = findBoxTarget(maze, N, M); while (!q.empty()) { St st = q.front(); q.pop(); for (St to_st : allPossibleOutcomes(maze, st, N, M)) { if (vis.find(to_st) == vis.end()) { q.push(to_st); steps.insert(MP(to_st, steps[st] + 1)); vis.insert(to_st); if (to_st.Y == target_box) { cout << steps[to_st] << '\n'; exit(0); } } } } cout << -1 << '\n'; return 0; }
#include <iostream> #include <stdlib.h> #include <memory.h> #include <utility> #include <queue> #include <stack> #include <set> #include <iterator> #include <algorithm> #include <limits.h> using namespace std; int N; int M; bool MAP[8][8]; int px, py; int tx, ty; int xx, xy; int V[64][64]; inline int amin(int a, int b, int c, int d) { int i=(a<b?a:b); int j=(c<d?c:d); return i<j?i:j; } inline bool inrange(int x, int y){ return x>=0 && x<N && y>=0 && y<M; } bool con(int x1, int y1, int x2, int y2, int * len){ if (x1==x2 && y1 == y2) {*len = 0; return true;} set<pair<int, int> > cur; set<pair<int, int> > tar; set<pair<int, int> > next; bool VISIT[8][8]; for(int i=0;i<8;i++) for(int j=0;j<8;j++) VISIT[i][j] = false; int cnt= 0; bool reach = false; cur.insert(make_pair(x1, y1)); tar.insert(make_pair(x2, y2)); while(!cur.empty() && !tar.empty()) { set<pair<int, int> > *phead; set<pair<int, int> > *ptail; if (cur.size() > tar.size()) {phead = &tar; ptail = &cur;} else {phead = &cur; ptail = &tar;} next.clear(); set<pair<int, int> >::iterator it; for(it=phead->begin();it!=phead->end();it++){ pair<int, int> pos = *it; if (ptail->find(pos)!=ptail->end()) {reach = true;break;} else { VISIT[pos.first][pos.second] = true; if (inrange(pos.first+1, pos.second) && !VISIT[pos.first+1][pos.second] && MAP[pos.first+1][pos.second]) {next.insert(make_pair(pos.first+1, pos.second));}// printf("\t-> <%d,%d>", pos.first+1, pos.second);} if (inrange(pos.first-1, pos.second) && !VISIT[pos.first-1][pos.second] && MAP[pos.first-1][pos.second]) {next.insert(make_pair(pos.first-1, pos.second));}// printf("\t-> <%d,%d>", pos.first-1, pos.second);} if (inrange(pos.first, pos.second+1) && !VISIT[pos.first][pos.second+1] && MAP[pos.first][pos.second+1]) {next.insert(make_pair(pos.first, pos.second+1));}// printf("\t-> <%d,%d>", pos.first, pos.second+1);} if (inrange(pos.first, pos.second-1) && !VISIT[pos.first][pos.second-1] && MAP[pos.first][pos.second-1]) {next.insert(make_pair(pos.first, pos.second-1));}// printf("\t-> <%d,%d>", pos.first, pos.second-1);} } } if (reach) break; else *phead = next; cnt+=1; } *len = cnt; return reach; } typedef pair<int, int> Pos; typedef pair<Pos, Pos> Pos_pair; struct State { Pos_pair pos; int len; friend bool operator < (const State &s1, const State &s2){ if (s1.len < s2.len) return true; else return false; } friend bool operator > (const State &s1, const State &s2){ if (s1.len > s2.len) return true; else return false; } }; int bfs(){ priority_queue<State, deque<State>, greater<State> > Q; // box<x,y> player<x,y> State init; init.pos = make_pair(make_pair(xx, xy), make_pair(px, py)); init.len = 0; Q.push(init); MAP[xx][xy] = true; while(!Q.empty()) { State s = Q.top(); Q.pop(); Pos xpos = s.pos.first; Pos ppos = s.pos.second; int x = xpos.first; int y = xpos.second; px = ppos.first; py = ppos.second; MAP[x][y]=false; if (V[x*8+y][px*8+py]==-2) V[x*8+y][px*8+py]=-1; if (x == tx && y == ty) {return s.len;} int len = 0; //down if (inrange(x+1, y) && inrange(x-1, y) && MAP[x+1][y] && MAP[x-1][y] && V[(x+1)*8+(y)][(x)*8+(y)]==-2 && con(px, py, x-1, y, &len)) { State next_state; next_state.pos = make_pair(make_pair(x+1, y), make_pair(x, y)); next_state.len = s.len + len+1; Q.push(next_state); } // up if (inrange(x-1, y) && inrange(x+1, y) && MAP[x-1][y] && MAP[x+1][y] && V[(x-1)*8+(y)][(x)*8+(y)]==-2 &&con(px, py, x+1, y, &len)) { State next_state; next_state.pos = make_pair(make_pair(x-1, y), make_pair(x, y)); next_state.len = s.len + len+1; Q.push(next_state); } // left if (inrange(x, y-1) && inrange(x, y+1) && MAP[x][y-1] && MAP[x][y+1] && V[(x)*8+(y-1)][(x)*8+(y)]==-2 && con(px, py, x, y+1, &len)) { State next_state; next_state.pos = make_pair(make_pair(x, y-1), make_pair(x, y)); next_state.len = s.len + len+1; Q.push(next_state); } // right if (inrange(x, y+1) && inrange(x, y-1) && MAP[x][y+1] && MAP[x][y-1] && V[(x)*8+(y+1)][(x)*8+(y)]==-2 && con(px, py, x, y-1, &len)) { State next_state; next_state.pos = make_pair(make_pair(x, y+1), make_pair(x, y)); next_state.len = s.len + len+1; Q.push(next_state); } MAP[xpos.first][xpos.second]=true; } return -1; } int main() { while(cin>>N){ cin>>M; if (N==0 || M==0) {cout<<0<<endl; continue;} for(int i=0;i<64;i++) for(int j=0;j<64;j++) V[i][j]=-2; string s; for(int i=0;i<N;i++){ cin>>s; for(int j=0;j<M;j++) { if (s[j] == '.') MAP[i][j] = true; if (s[j] == '#') MAP[i][j] = false; if (s[j] == 'X') {MAP[i][j] = true; px=i;py=j;} if (s[j] == '@') {MAP[i][j] = true; tx=i;ty=j;} if (s[j] == '*') {MAP[i][j] = false; xx=i;xy=j;} } } cout<<bfs()<<endl; } return 0; }