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HDU 1401 Solitaire

編輯：關於C++

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 5
2 4 3 3 3 6 4 6

Sample Output

YES

Source Southwestern Europe 2002

雙向廣搜。每組棋子的坐標為1~8 共有四個坐標 8個數都可以用三個二進制數保存所以可以將狀態轉換為相應的十進制數進行狀態壓縮

只能走八步，每步有十六中變化所以雙向廣搜可以解決

#include
#include
#include
#include
#include
#include
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];  //標記  狀態壓縮
int us[65536+10];  //記錄出現過的狀態
int len;

bool cmp(Point n1,Point n2)
{
    return n1.x!=n2.x?n1.xus[mid])
            l=mid+1;
        else r=mid-1;
    }
    return false ;
}

void dfs()
{
    queueq;
    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;
    //起點開始廣搜可以出現的狀態
    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))  //從起點走出現過這種狀態
    {
        printf(YES
);
        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
);
                        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
);
                        return ;
                    }
                    if(!vis[Hash])
                    {
                        vis[Hash]=true ;
                        q.push(e);
                    }
                }
            }
        }
    }
    printf(NO
);
    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;
        //將坐標轉為0~7對應2^3-1的數  狀態壓縮
        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;
}