poj 2965 The Pilots Brothers' refrigerator

//位压缩加搜索枚举，用pre记录其前驱



#include<stdio.h>

#include<string.h>

#define N 100000

struct node

{

    int x,y,num,pre,step;

}p[N*4];

struct nn

{

    int x,y;

}g[N*4];

char str[10][10];

int vis[N];

int d[16]=

{

    0xf888,0xf444,0xf222,0xf111,

    0x8f88,0x4f44,0x2f22,0x1f11,

    0x88f8,0x44f4,0x22f2,0x11f1,

    0x888f,0x444f,0x222f,0x111f,

};



int bfs(int sum)

{

    int head=0,tail=0,i;

    p[head].num=sum;

    p[head].step=0;

    p[head].pre=0;

    tail++;

    while(head<tail)

    {

        int k=p[head].num;

        for(i=0;i<16;i++)

        {

            int x=k^d[i];



            if(vis[x]==0)

            {

                vis[x]=1;



                p[tail].num=x;

                p[tail].x=i/4;

                p[tail].y=i%4;

                p[tail].pre=head;

                p[tail].step=p[head].step+1;

                 if(x==0){  return tail;}

                 tail++;



            }

        }

        head++;

    }

    return 0;





}

int main()

{

    int i,j;

    memset(vis,0,sizeof(vis));

    for(i=0;i<4;i++)

    {

        scanf("%s",str[i]);

    }

    int sum=0;

    for(i=0;i<4;i++)

    {

        for(j=0;j<4;j++)

        {

            if(str[i][j]=='+')

            {

                sum<<=1;

                sum+=1;

            }

            else

            {

                sum<<=1;

            }

        }

    }



    if(sum==0){printf("0\n");return 0;}

    int ans=bfs(sum);



    printf("%d\n",p[ans].step);

    i=0;

    g[i].x=p[ans].x;

    g[i].y=p[ans].y;//注意这，可能只需一次就可以开锁

    while(p[ans].pre)

    {



        //printf("%d %d\n",p[ans].x,p[ans].y);

        i++;

        ans=p[ans].pre;

        if(ans==0)break;

        g[i].x=p[ans].x;

        g[i].y=p[ans].y;

    }



    for(;i>=0;i--)

    {

        printf("%d %d\n",g[i].x+1,g[i].y+1);

    }





}