Holedox Moving POJ

题目链接：Holedox Moving POJ - 1324

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Holedox Moving

Time Limit: 5000 ms
Memory Limit: 65536 kB

Description

During winter, the most hungry and severe time, Holedox sleeps in its lair. When spring comes, Holedox wakes up, moves to the exit of its lair, comes out, and begins its new life.
Holedox is a special snake, but its body is not very long. Its lair is like a maze and can be imagined as a rectangle with n*m squares. Each square is either a stone or a vacant place, and only vacant places allow Holedox to move in. Using ordered pair of row and column number of the lair, the square of exit located at (1,1).

Holedox’s body, whose length is L, can be represented block by block. And let B1(r1,c1) B2(r2,c2) … BL(rL,cL) denote its L length body, where Bi is adjacent to Bi+1 in the lair for 1 <= i <=L-1, and B1 is its head, BL is its tail.

To move in the lair, Holedox chooses an adjacent vacant square of its head, which is neither a stone nor occupied by its body. Then it moves the head into the vacant square, and at the same time, each other block of its body is moved into the square occupied by the corresponding previous block.

For example, in the Figure 2, at the beginning the body of Holedox can be represented as B1(4,1) B2(4,2) B3(3,2)B4(3,1). During the next step, observing that B1’(5,1) is the only square that the head can be moved into, Holedox moves its head into B1’(5,1), then moves B2 into B1, B3 into B2, and B4 into B3. Thus after one step, the body of Holedox locates in B1(5,1)B2(4,1)B3(4,2) B4(3,2) (see the Figure 3).

Given the map of the lair and the original location of each block of Holedox’s body, your task is to write a program to tell the minimal number of steps that Holedox has to take to move its head to reach the square of exit (1,1).

Input

The input consists of several test cases. The first line of each case contains three integers n, m (1<=n, m<=20) and L (2<=L<=8), representing the number of rows in the lair, the number of columns in the lair and the body length of Holedox, respectively. The next L lines contain a pair of row and column number each, indicating the original position of each block of Holedox’s body, from B1(r1,c1) to BL(rL,cL) orderly, where 1<=ri<=n, and 1<=ci<=m,1<=i<=L. The next line contains an integer K, representing the number of squares of stones in the lair. The following K lines contain a pair of row and column number each, indicating the location of each square of stone. Then a blank line follows to separate the cases.

The input is terminated by a line with three zeros.

Note: Bi is always adjacent to Bi+1 (1<=i<=L-1) and exit square (1,1) will never be a stone.

Output

For each test case output one line containing the test case number followed by the minimal number of steps Holedox has to take. “-1” means no solution for that case.

Sample Input

5 6 4
4 1
4 2
3 2
3 1
3
2 3
3 3
3 4

4 4 4
2 3
1 3
1 4
2 4
4

2 1
2 2
3 4
4 2

0 0 0

Sample Output

Case 1: 9
Case 2: -1

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题意：一个n,m的巢穴地图，里面有石头不能走，洞口永远为（1,1）。然后给你一条长度L[2,8]的蛇。然后蛇每一次移动，身体也会跟着走，然后你不能穿过石头和身体。问你到洞口的最短步数是多少？

算法：A*+状态压缩

思路：

这里启发函数就直接蛇头与洞口曼哈顿距离就行了；
判重函数：这里每个点维护的信息：蛇头坐标，步数，蛇身体状态；其实区分每一个点入队的判重，这里就重要是蛇头和身体状态（因为身体也会阻碍蛇头前进）；蛇头直接二维x,y就行了；然后就是蛇身体如何区分记录呢？一开始我的想法是，最大长度为8，那再记录身体其他坐标就好了，但是21的16次方直接爆了，MLE。然后想了很久，还是看了别人博客，学到了。只要记录蛇头，然后第二节只要记录方向就行，第三节只要记录相对于第二节的方向就行了，等等。因为方向只有4个方向，所以最大只要 4的7次方就是 2 的14次方；所以开三维，vis[21][21][1<<14]

难点：

我直接1A了，没有什么地方出错；求方向四进制状态、还原身体坐标、维护新的四进制状态这几个可能会比较容易搞得混乱，我自己也码的不好看，但也懒得改了，希望能帮到你

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#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cstdio>using namespace std;const int INF = 0x3f3f3f3f;
const int MAXN = 1010;int four[7];//四进制基数
int n,m;//迷宫长宽
int stNum;bool ditu[21][21];//石头数+地图标记
int snakeNum;//蛇的长度
int snakeHeadX,snakeHeadY,snakeBody;//蛇头坐标  + 蛇形状的四进制表示（用四进制记录每一节对于前一节的方向）
bool vis[21][21][1<<14];//蛇头坐标  + 蛇形状的四进制表示 de  三维标记判重函数
int xx[] = {0,1,0,-1};
int yy[] = {1,0,-1,0};//用于求蛇身每一节对于前一节的方向，0,1,2,3，分别表示右下左上
int check(int a,int b,int c,int d){if(c-a==0&&d-b==1) return 0;if(c-a==1&&d-b==0) return 1;if(c-a==0&&d-b==-1) return 2;if(c-a==-1&&d-b==0) return 3;
}//初始输入+预处理
void init(){cin>>snakeHeadX>>snakeHeadY;int a=snakeHeadX,b=snakeHeadY,c,d;snakeBody=0;for(int i=0;i<snakeNum-1;i++){cin>>c>>d;snakeBody+=four[i]*check(a,b,c,d);a=c,b=d;}cin>>stNum;memset(ditu,true,sizeof ditu);for(int i=0;i<stNum;i++){cin>>a>>b;ditu[a][b]=false;}
}//测试函数 可删除
void show(){cout<<n<<" "<<m<<" "<<snakeNum<<endl;cout<<snakeHeadX<<" "<<snakeHeadY<<" "<<snakeBody<<endl;for(int i=0;i<snakeNum-1;i++){int tmp = snakeBody % 4;cout<<tmp<<" ";snakeBody/=4;}cout<<endl;cout<<stNum<<endl;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++) cout<<ditu[i][j]<<" ";cout<<endl;}
}//蛇头X,Y坐标 蛇的四进制表示  启发函数数值f（这里是蛇头与洞口（1,1）的曼哈顿距离）
struct node{int x,y,snake,step,f;node(int xx,int yy,int ssnake,int sstep):x(xx),y(yy),snake(ssnake),step(sstep){f=step+abs(x-1)+abs(y-1);}bool operator < (const node &a) const{return f > a.f;}
};//这个函数是通过 蛇头坐标 蛇身四进制 来求 蛇身坐标数组 和 蛇尾对于前一节的方向
void snakeXY(int SX,int SY,int (&snakeX)[8],int (&snakeY)[8],int &c,int d){int x = SX,y = SY;for(int i=0;i<snakeNum-1;i++){int tmp = d % 4; d/=4;x+=xx[tmp],y+=yy[tmp];snakeX[i] = x ,snakeY[i] = y ;if(i==snakeNum-2) c = tmp;}
}//判断蛇头是否能走(x,y)
bool checkN(int x,int y,int a[],int b[]){if(x<=0||x>n||y<=0||y>m||ditu[x][y]==false) return false;for(int i=0;i<snakeNum-1;i++){if(x==a[i]&&y==b[i]) return false;}return true;
}//A*算法
int Astar(){memset(vis,false,sizeof vis);priority_queue<node> q;q.push(node(snakeHeadX,snakeHeadY,snakeBody,0));vis[snakeHeadX][snakeHeadY][snakeBody] = true;while(!q.empty()){node p = q.top();q.pop();if(p.x==1&&p.y==1) return p.step;//这里记录蛇身坐标 和 蛇尾方向int snakeX[8],snakeY[8],tail;snakeXY(p.x,p.y,snakeX,snakeY,tail,p.snake);//右、下、左、上for(int i=0;i<4;i++){int x = p.x + xx[i],y = p.y + yy[i];if(!checkN(x,y,snakeX,snakeY)) continue;//这里开始算新的方向四进制  去尾加头int body = p.snake - four[snakeNum-2]*tail;body*=4;if(i==0) body+=2;else if(i==1) body+=3;else if(i==2) body+=0;else body += 1;if(vis[x][y][body]) continue;vis[x][y][body] = true;q.push(node(x,y,body,p.step+1));}}return -1;
}int main()
{four[0]=1;for(int i=1;i<7;i++) four[i] = four[i-1]*4;int _=1;while(cin>>n>>m>>snakeNum&&n&&m&&snakeNum){init();cout<<"Case "<<_++<<": ";cout<<Astar()<<endl;}return 0;
}