Bloxorz I
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 2624 |
|
Accepted: 882 |
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.
Sample Input
7 7
#######
#..X###
#..##O#
#....E#
#....E#
#.....#
#######
0 0
Sample Output
10
Source
此題是標(biāo)準(zhǔn)的BFS,但是要注意優(yōu)化，寫的不好代碼很容易超過5000B.
1.首先是狀態(tài)壓縮，積木可以立起來，也可以朝向上下左右躺下，其實向左躺和向右躺下實際上是同一種狀態(tài)，向上躺和向下躺亦然，所以本題可以壓縮成3個狀態(tài)，這里用1表示直立，2表示橫躺，3表示豎躺，可以節(jié)省大量空間花費；
2.用隊列進行廣搜的時候最好不要使用STL的queue容器，由于queue的初始值有限，搜索的時候需要大量移動內(nèi)存中的數(shù)據(jù)，效率比較低，不妨手工直接模擬一個隊列。
3.在搜索的時候，由于狀態(tài)空間比較大，如果一一列出，代碼過于冗長，比較好的方法是用一個三維矩陣來存儲狀態(tài)之間的關(guān)系，這樣搜索的過程可以只用一個for循環(huán)來實現(xiàn)，代碼十分精簡。
4.最好進行初始化，讀入數(shù)據(jù)的時候?qū)⑵滢D(zhuǎn)化成整數(shù)，這樣操作起來比較方便。
下面是我的代碼：
//This is the sorce code for POJ 3322
//created by abilitytao
//Time:2009年8月16日22:33:08
//special thanks to Rainer
#include<iostream>
using namespace std;
#define MAX 502
#define LINEMAX 9000000
struct node
{
int x,y;
int step;
int state;//1代表直立，2代表水平放，3代表垂直放
};
bool visit[MAX][MAX][4];
int graph[MAX][MAX];//0->empty cell，1->weak cell,2->rigid cell
node l[LINEMAX];
int dir[3][4][3]={{{0,-1,2},{-2,0,1},{0,2,2},{1,0,1}},
{{0,-1,0},{-1,0,-1},{0,1,0},{2,0,-1}},
{{0,-2,-2},{-1,0,0},{0,1,-2},{1,0,0}}};
inline bool check(int x,int y,int state)
{
if(graph[x][y]==0) return false;
if(state==1&&graph[x][y]==1) return false;
else if(state==2&&graph[x+1][y]==0) return false;
else if(state==3&&graph[x][y-1]==0) return false;
return true;
}
inline node inputgraph(int r,int c,int &endx,int &endy)
{
int i,j;
int len;
char temp[MAX];
int x1=0,y1=0;
int x2=0,y2=0;
int flag=0;
node s;
memset(graph,0,sizeof(graph));//Graph矩陣清0
memset(visit,0,sizeof(visit));//visit矩陣清0
for(i=1;i<=r;i++)
{
scanf("%s",temp);
len=strlen(temp);
for(j=0;j<len;j++)
{
if(temp[j]=='#') graph[j+1][i]=0;
else if(temp[j]=='E') graph[j+1][i]=1;
else if(temp[j]=='.') graph[j+1][i]=2;
else if(temp[j]=='X'&&flag==0)
{
graph[j+1][i]=2;
x1=j+1;y1=i;
flag=1;
}
else if(temp[j]=='X'&&flag==1)
{
graph[j+1][i]=2;
x2=j+1;y2=i;
}
else if(temp[j]=='O')
{
graph[j+1][i]=2;
endx=j+1;
endy=i;
}
}
}
if(x1!=0&&x2!=0&&y1!=0&&y2!=0)
{
if(y1<y2) {s.x=x1;s.y=y2;s.state=3;}
else if(x1<x2) {s.x=x1;s.y=y1;s.state=2;}
}
else {s.x=x1;s.y=y1;s.state=1;}
s.step=0;
return s;
}
int main()
{
int r,c;
int i,j;
int endx,endy;
int newx,newy;
int newstate;
while(scanf("%d%d",&r,&c))
{
if(r==0&&c==0)
break;
int front=1;
int rear=1;
l[front]=inputgraph(r,c,endx,endy);
visit[l[front].x][l[front].y][l[front].state]=1;
while(front<=rear)
{
if(l[front].x==endx&&l[front].y==endy&&l[front].state==1)
{
printf("%d\n",l[front].step);
break;
}
for(j=0;j<4;j++)
{
newx=l[front].x+dir[l[front].state-1][j][0];
newy=l[front].y+dir[l[front].state-1][j][1];
newstate=l[front].state+dir[l[front].state-1][j][2];
if(!visit[newx][newy][newstate]&&check(newx,newy,newstate))
{
rear=rear++;
l[rear].x=newx;
l[rear].y=newy;
l[rear].state=newstate;
visit[newx][newy][newstate]=1;
l[rear].step=l[front].step+1;
}
}
front++;
}
if(front>rear)
printf("Impossible\n");
}
return 0;
}