poj3322 Bloxorz I

Description

Little Tom loves playing games. One day he downloads a little computer
game called ‘Bloxorz’ which makes him excited. It’s a game about
rolling a box to a specific position on a special plane. Precisely,
the plane, which is composed of several unit cells, is a rectangle
shaped area. And the box, consisting of two perfectly aligned unit
cube, may either lies down and occupies two neighbouring cells or
stands up and occupies one single cell. One may move the box by
picking one of the four edges of the box on the ground and rolling the
box 90 degrees around that edge, which is counted as one move. There
are three kinds of cells, rigid cells, easily broken cells and empty
cells. A rigid cell can support full weight of the box, so it can be
either one of the two cells that the box lies on or the cell that the
box fully stands on. A easily broken cells can only support half the
weight of the box, so it cannot be the only cell that the box stands
on. An empty cell cannot support anything, so there cannot be any part
of the box on that cell. The target of the game is to roll the box
standing onto the only target cell on the plane with minimum moves.

The box stands on a single cell

The box lies on two neighbouring cells, horizontally

The box lies on two neighbouring cells, vertically

After Little Tom passes several stages of the game, he finds it much
harder than he expected. So he turns to your help.

Input

Input contains multiple test cases. Each test case is one single stage
of the game. It starts with two integers R and C(3 ≤ R, C ≤ 500) which
stands for number of rows and columns of the plane. That follows the
plane, which contains R lines and C characters for each line, with ‘O’
(Oh) for target cell, ‘X’ for initial position of the box, ‘.’ for a
rigid cell, ‘#’ for a empty cell and ‘E’ for a easily broken cell. A
test cases starts with two zeros ends the input.

It guarantees that
There's only one 'O' in a plane.
There's either one 'X' or neighbouring two 'X's in a plane.
The first(and last) row(and column) must be '#'(empty cell).
Cells covered by 'O' and 'X' are all rigid cells.
Output

For each test cases output one line with the minimum number of moves
or “Impossible” (without quote) when there’s no way to achieve the
target cell.

bfs暴搜。保存状态的时候，记录摆法（三种）和左上角方块的位置，用打表数组转移。

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int map[510][510],m,n,kk[3][4]={{2,2,1,1},{1,1,0,0},{0,0,2,2}};
int xx[3][4]={{1,-2,0,0},{1,-1,0,0},{2,-1,0,0}};
int yy[3][4]={{0,0,-2,1},{0,0,-1,2},{0,0,-1,1}};
bool vis[510][510][3];
struct con
{int x,y,k,t;bool ok(){if (vis[x][y][k]) return 0;if (k==0)return x>=1&&x<=n&&y>=1&&y<=m&&(!map[x][y]);if (k==1)return x>=1&&x<=n&&y>=1&&y<m&&map[x][y]<2&&map[x][y+1]<2;return x>=1&&x<n&&y>=1&&y<=m&&map[x][y]<2&&map[x+1][y]<2;}bool meet(con ccc){return x==ccc.x&&y==ccc.y&&k==ccc.k;}
}c1,c2,tar;
queue<con> q;
char s[510];
int main()
{int i,j,k,p,x,y,z,w,cnt;bool flag;while (scanf("%d%d",&n,&m)&&n&&m){memset(vis,0,sizeof(vis));while (!q.empty()) q.pop();cnt=0;for (i=1;i<=n;i++){scanf("%s",s+1);for (j=1;j<=m;j++)if (s[j]=='#') map[i][j]=2;else{if (s[j]=='E') map[i][j]=1;else{map[i][j]=0;if (s[j]=='O') tar=(con){i,j,0,0};if (s[j]=='X'){cnt++;if (cnt==1) x=i,y=j;else z=i,w=j;}}}}if (cnt==1) q.push((con){x,y,0,0}),vis[x][y][0]=1;else{if (x==z) q.push((con){x,y,1,0}),vis[x][y][1]=1;else q.push((con){x,y,2,0}),vis[x][y][2]=1;}flag=0;while (!q.empty()){c1=q.front();q.pop();for (i=0;i<4;i++){c2=(con){c1.x+xx[c1.k][i],c1.y+yy[c1.k][i],kk[c1.k][i],c1.t+1};if (!c2.ok()) continue;if (c2.meet(tar)){flag=1;printf("%d\n",c2.t);break;}q.push(c2);vis[c2.x][c2.y][c2.k]=1;}if (flag) break;}if (!flag) printf("Impossible\n");}
}

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