POJ-1324-Holedox Moving(BFS)

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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

Output

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

Hint

Source

思路：从头部开始想尾部走，不同方向代表不同的值，就得到蛇的状态，然后就是普通的BFS，还要注意不能咬到自己，首尾相连也不行。

#include <stdio.h>

struct S{
int step,x[9],y[9],head;
}que[1000000];

bool vis[20][20][16384];
int mp[20][20],nxt[4][2]={{1,0},{0,-1},{0,1},{-1,0}},mi[8]={1,4,16,64,156,1024,4096,16384};

int main()
{
    int n,m,l,i,j,k,x,y,head,temp,lx,ly,tl,valx,valy,casenum=1,top,bottom;

    while(~scanf("%d%d%d",&n,&m,&l) && n)
    {
        for(i=0;i<l;i++)
        {
            scanf("%d%d",&x,&y);

            que[0].x[i]=x-1;
            que[0].y[i]=y-1;
        }

        for(i=0;i<n;i++) for(j=0;j<m;j++) mp[i][j]=1;
        for(i=0;i<n;i++) for(j=0;j<m;j++) for(k=0;k<mi[l-1];k++) vis[i][j][k]=0;

        scanf("%d",&k);

        while(k--)
        {
            scanf("%d%d",&x,&y);

            mp[x-1][y-1]=0;
        }

        que[0].step=0;
        que[0].head=0;

        top=0;
        bottom=1;

        while(top<bottom)
        {
            if(!que[top].x[que[top].head] && !que[top].y[que[top].head])
            {
                printf("Case %d: %d\n",casenum++,que[top].step);

                break;
            }

            head=que[top].head-1;

            if(head==-1) head=l;

            que[top].step++;

            for(i=0;i<4;i++)
            {
                x=que[top].x[head]=que[top].x[que[top].head]+nxt[i][0];
                y=que[top].y[head]=que[top].y[que[top].head]+nxt[i][1];

                if(x>=0 && x<n && y>=0 && y<m && mp[x][y])
                {
                    temp=0;

                    lx=x;
                    ly=y;

                    tl=0;

                    for(j=que[top].head;tl<l;j++)
                    {
                        if(j==l+1) j=0;

                        if(que[top].x[j]==x && que[top].y[j]==y) break;//如果会咬到自己

                        if(tl<l-1)//计算状态
                        {
                            valx=lx-que[top].x[j];
                            valy=ly-que[top].y[j];

                            if(valx==0 && valy==1) temp=temp<<2;
                            else if(valx==1 && valy==0) temp=temp<<2|1;
                            else if(valx==-1 && valy==0) temp=temp<<2|2;
                            else if(valx==0 && valy==-1) temp=temp<<2|3;

                            lx=que[top].x[j];
                            ly=que[top].y[j];
                        }

                        tl++;
                    }

                    if(tl==l && !vis[x][y][temp])//如果不会咬到自己并且该状态没有访问过
                    {
                        vis[x][y][temp]=1;

                        que[top].head--;

                        if(que[top].head==-1) que[top].head=l;

                        que[top].x[head]=x;
                        que[top].y[head]=y;

                        que[bottom++]=que[top];

                        que[top].head++;

                        if(que[top].head==l+1) que[top].head=0;
                    }
                }
            }

            top++;
        }

        if(top==bottom) printf("Case %d: -1\n",casenum++);
    }
}

POJ-1324-Holedox Moving(BFS),布布扣,bubuko.com

POJ-1324-Holedox Moving(BFS)

标签：des   style   blog   http   color   os   

原文地址：http://blog.csdn.net/faithdmc/article/details/37296101