山寨版的双向BFS。。。

正着搜的结果存下来，反着收的结果在正着搜的结果里找。。。

Solitaire

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.

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

YES

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>

using namespace std;

const int dir_x[4]={-1,1,0,0};
const int dir_y[4]={0,0,-1,1};

bool inside(int x,int y)
{
    if((x>=1&&x<=8)&&(y>=1&&y<=8)) return true;
    return false;
}

struct POINT
{
    int x,y;
    bool operator<(const POINT& res) const
    {
        if(x!=res.x) return x<res.x;
        return y<res.y;
    }
};

POINT st[4],ed[4];
int hahashsh(POINT st[])
{
    int haha=0;
    for(int i=0;i<4;i++)
    {
        haha=haha*10+st[i].x;
        haha=haha*10+st[i].y;
    }
    return haha;
}

set<int> panchong;

struct NODE
{
    int id,ti;
};

int doit(int xx,int yy,int hs)
{
    POINT rk[4];
    int u=hs;
    for(int i=3;i>=0;i--)
    {
        rk[i].y=u%10; u/=10;
        rk[i].x=u%10; u/=10;
    }

    int x=rk[xx].x,y=rk[xx].y;
    int X=x+dir_x[yy],Y=y+dir_y[yy];

    if(inside(X,Y)==false) return -1;

    bool flag=false;
    for(int i=0;i<4;i++) if(rk[i].x==X&&rk[i].y==Y) { flag=true; break; }

    if(flag==true)
    {
        int X2=X+dir_x[yy],Y2=Y+dir_y[yy];
        if(inside(X2,Y2)==false) return -1;
        bool flag2=false;
        for(int i=0;i<4;i++) if(rk[i].x==X2&&rk[i].y==Y2) { flag2=true; break; }
        if(flag2==true) return -1;
        rk[xx].x=X2;rk[xx].y=Y2;
    }
    else
    {
        rk[xx].x=X;rk[xx].y=Y;
    }

    sort(rk,rk+4);

    return hahashsh(rk);
}

void zBFS(int ss)
{
    queue<NODE> q;
    q.push((NODE){ss,0});
    while(!q.empty())
    {
        NODE u,v;
        u=q.front();q.pop();
        if(u.ti+1>4) continue;
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<4;j++)
            {
                ///4个棋子4个方向
                int temp=doit(i,j,u.id);
                if(temp==-1) continue;
                if(panchong.count(temp)==0)
                {
                    panchong.insert(temp);
                    q.push((NODE){temp,u.ti+1});
                }
            }
        }
    }
}

int fBFS(int es)
{
    queue<NODE> q;
    set<int> fafa;
    fafa.clear();
    fafa.insert(es);
    q.push((NODE){es,0});
    while(!q.empty())
    {
        NODE u,v;
        u=q.front(); q.pop();
        if(u.ti+1>4) continue;
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<4;j++)
            {
                 ///4个棋子4个方向
                int temp=doit(i,j,u.id);
                if(temp==-1) continue;
                if(panchong.count(temp)) return 1;
                if(fafa.count(temp)==0)
                {
                    fafa.insert(temp);
                    q.push((NODE){temp,u.ti+1});
                }
            }
        }
    }
    return 0;
}

int solve(int ss,int es)
{
    zBFS(ss);
    return fBFS(es);
}

int main()
{
while(scanf("%d%d",&st[0].x,&st[0].y)!=EOF)
{
    for(int i=1;i<4;i++) scanf("%d%d",&st[i].x,&st[i].y);
    for(int i=0;i<4;i++) scanf("%d%d",&ed[i].x,&ed[i].y);
    sort(st,st+4);sort(ed,ed+4);
    panchong.clear();
    panchong.insert(hahashsh(st));

    if(solve(hahashsh(st),hahashsh(ed))) puts("YES"); else puts("NO");
}
    return 0;
}

HDOJ 1401 Solitaire

原文：http://blog.csdn.net/u012797220/article/details/19361691