HDU 1401 Solitaire

Solitaire

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 12   Accepted Submission(s) : 1

Problem Description

Solitaire is a game played on a chessboard 8x8. The rows and columns of the chessboard are numbered from 1 to 8, from the top to the bottom and from left to right respectively.

There are four identical pieces on the board. In one move it is allowed to:

> move a piece to an empty neighboring field (up, down, left or right),

> jump over one neighboring piece to an empty field (up, down, left or right). 



There are 4 moves allowed for each piece in the configuration shown above. As an example let's consider a piece placed in the row 4, column 4. It can be moved one row up, two rows down, one column left or two columns right.

Write a program that:

> reads two chessboard configurations from the standard input,

> verifies whether the second one is reachable from the first one in at most 8 moves,

> writes the result to the standard output.

 

Input

Each of two input lines contains 8 integers a1, a2, ..., a8 separated by single spaces and describes one configuration of pieces on the chessboard. Integers a2j-1 and a2j (1 <= j <= 4) describe the position of one piece - the row number and the column number respectively. Process to the end of file.

 

Output

The output should contain one word for each test case - YES if a configuration described in the second input line is reachable from the configuration described in the first input line in at most 8 moves, or one word NO otherwise.

 

Sample Input

4 4 4 5 5 4 6 52 4 3 3 3 6 4 6

 

Sample Output

YES

 

Source

Southwestern Europe 2002

 

双向广搜。  每组棋子的坐标为1~8  共有四个坐标  8个数都可以用三个二进制数保存    所以可以将状态转换为相应的十进制数进行状态压缩

只能走八步，每步有十六中变化  所以双向广搜可以解决

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<map>#include<algorithm>using namespace std;struct Point{    int x,y;    bool cheak()    {        if(x>=0&&x<8&&y>=0&&y<8)            return true ;        else return false ;    }};struct node{    Point a[5];    int t;} st,ed,e;int dir[4][2]= {0,1,0,-1,1,0,-1,0};bool vis[16777216+1];  //标记  状态压缩int us[65536+10];  //记录出现过的状态int len;bool cmp(Point n1,Point n2){    return n1.x!=n2.x?n1.x<n2.x:n1.y<n2.y;}int get_hash(Point *tmp)  //状态压缩{    int res=0;    sort(tmp,tmp+4,cmp);    for(int i=0; i<4; i++)    {        res|=(tmp[i].x<<(6*i));        res|=(tmp[i].y<<(6*i+3));    }    return res;}bool fine(int num)   //二分查找{    int l=0,r=len-1;    while(l<=r)    {        int mid=(l+r)/2;        if(num==us[mid]) return true ;        if(num>us[mid])            l=mid+1;        else r=mid-1;    }    return false ;}void dfs(){    queue<node>q;    memset(vis,false ,sizeof(vis));    len=0;    int Hash;    Hash=get_hash(st.a);    vis[Hash]=true ;    us[len++]=Hash;    st.t=0;    q.push(st);    int k;    //起点开始广搜可以出现的状态    while(q.size())    {        st=q.front();        q.pop();        if(st.t>=4) continue ;        for(int j=0; j<4; j++)        {            for(int i=0; i<4; i++)            {                e=st;                e.t++;                e.a[i].x+=dir[j][0];                e.a[i].y+=dir[j][1];                if(!e.a[i].cheak())                    continue ;                for(k=0; k<4; k++)                {                    if(k!=i&&e.a[k].x==e.a[i].x&&e.a[k].y==e.a[i].y)                        break;                }                if(k==4)                {                    Hash=get_hash(e.a);                    if(!vis[Hash])                    {                        vis[Hash]=true ;                        us[len++]=Hash;                        q.push(e);                    }                }                else                {                    e.a[i].x+=dir[j][0];                    e.a[i].y+=dir[j][1];                    if(!e.a[i].cheak())                        continue ;                    for(k=0; k<4; k++)                    {                        if(k!=i&&e.a[k].x==e.a[i].x&&e.a[k].y==e.a[i].y)                            break;                    }                    if(k==4)                    {                        Hash=get_hash(e.a);                        if(!vis[Hash])                        {                            vis[Hash]=true ;                            us[len++]=Hash;                            q.push(e);                        }                    }                }            }        }    }    sort(us,us+len);    memset(vis,false ,sizeof(vis));    Hash=get_hash(ed.a);    if(fine(Hash))  //从起点走出现过这种状态    {        printf("YES\n");        return ;    }    vis[Hash]=true ;    q.push(ed);    while(q.size())    {        st=q.front();        q.pop();        if(st.t>=4) continue ;        for(int j=0; j<4; j++)        {            for(int i=0; i<4; i++)            {                e=st;                e.t++;                e.a[i].x+=dir[j][0];                e.a[i].y+=dir[j][1];                if(!e.a[i].cheak())                    continue ;                for(k=0; k<4; k++)                {                    if(k!=i&&e.a[k].x==e.a[i].x&&e.a[k].y==e.a[i].y)                        break;                }                if(k==4)                {                    Hash=get_hash(e.a);                    if(fine(Hash))                    {                        printf("YES\n");                        return ;                    }                    if(!vis[Hash])                    {                        vis[Hash]=true ;                        q.push(e);                    }                }                else                {                    e.a[i].x+=dir[j][0];                    e.a[i].y+=dir[j][1];                    if(!e.a[i].cheak()) continue ;                    for(k=0; k<4; k++)                    {                        if(k!=i&&e.a[k].x==e.a[i].x&&e.a[k].y==e.a[i].y)                            break;                    }                    if(k!=4) continue ;                    Hash=get_hash(e.a);                    if(fine(Hash))                    {                        printf("YES\n");                        return ;                    }                    if(!vis[Hash])                    {                        vis[Hash]=true ;                        q.push(e);                    }                }            }        }    }    printf("NO\n");    return ;}int main(){    while(~scanf("%d%d",&st.a[0].x,&st.a[0].y))    {        for(int i=1; i<4; i++)            scanf("%d%d",&st.a[i].x,&st.a[i].y);        st.t=0;        for(int i=0; i<4; i++)            scanf("%d%d",&ed.a[i].x,&ed.a[i].y);        ed.t=0;        //将坐标转为0~7对应2^3-1的数  状态压缩        for(int i=0; i<4; i++)        {            st.a[i].x--;              st.a[i].y--;            ed.a[i].x--;            ed.a[i].y--;        }        dfs();    }    return 0;}