hdu1401 Solitaire-优快云博客

本文链接：https://blog.youkuaiyun.com/henryascend/article/details/19476657

Solitaire

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2960 Accepted Submission(s): 920

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

Recommend

Ignatius.L

双向BFS，由于题目限定的步数，所以可以先做起点，再做终点，这样比较方便

但是对于其它题目，可以起点终点同时拓展

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

using namespace std;

struct data
{
    int s,idx;
};

struct Point{
    int x,y;
};


Point St[4],En[4],tmp[4];
int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};
bool bz[11][11];

set   <int>  s1, s2;
queue <data> q;

bool cmp(Point a,Point b){
    return (a.x<b.x|| (a.x==b.x&&a.y<b.y) );
}

int change( Point *a)
{
    sort(a,a+4,cmp);
    int t=0;
    for (int i=0; i<4; i++)
    {
        t=t*10+a[i].x;
        t=t*10+a[i].y;
    }
    return t;
}

void recovery( int x )
{
    for (int i=0; i<4; i++)
    {
        tmp[i].y=x % 10;
        x/=10;
        tmp[i].x=x % 10;
        x/=10;
    }
    sort(tmp,tmp+4,cmp);
}

bool bound(int tx,int ty)
{

        if (tx<1||tx>8||ty<1||ty>8 ) return false;
        return true;
}

int main()
{
    while (scanf("%d%d%d%d%d%d%d%d",&St[0].x,&St[0].y,&St[1].x,&St[1].y,&St[2].x,&St[2].y,&St[3].x,&St[3].y)!=EOF)
    {
        while (!q.empty()) q.pop();
        s1.clear();
        s2.clear();
        scanf("%d%d%d%d%d%d%d%d",&En[0].x,&En[0].y,&En[1].x,&En[1].y,&En[2].x,&En[2].y,&En[3].x,&En[3].y);
        sort(St,St+4,cmp);
        sort(En,En+4,cmp);
        int st=0, en=0;
        st=change(St);
        s1.insert(st);
        q.push( (data){st,0} );
        while (!q.empty())
        {
            data u=q.front();
            q.pop();
            for (int i=0; i<4; i++)
            {
                for (int j=0; j<4; j++)
                {
                    recovery( u.s );
                    memset(bz,0,sizeof(bz));
                    for (int k=0; k<4; k++)
                        bz[tmp[k].x][tmp[k].y]=1;

                    int tx=tmp[j].x+dx[i];
                    int ty=tmp[j].y+dy[i];
                    if (!bound(tx,ty))  continue;
                    if (bz[tx][ty] )
                    {
                        tx=tx+dx[i];
                        ty=ty+dy[i];
                    }
                    if (!bound(tx,ty))  continue;
                    if (bz[tx][ty])  continue;
                    tmp[j].x=tx; tmp[j].y=ty;

                    data v;
                    v.s=change( tmp );
                    if ( s1.count(v.s) ) continue;
                    s1.insert( v.s );

                    v.idx=u.idx+1;
                    if (v.idx<4)
                    {
                        q.push( v );
                    }
                }
            }
        }
        while (!q.empty()) q.pop();
        bool flag=0;
        en=change(En);
        if ( s1.count(en) )
        {
            printf("YES\n");
            continue;
        }
        s2.insert(en);
        q.push((data){en,0});
        while (!q.empty())
        {
            data u=q.front();
            q.pop();
            for (int i=0; i<4; i++)
            {
                for (int j=0; j<4; j++)
                {
                    recovery( u.s );
                    memset(bz,0,sizeof(bz));
                    for (int k=0; k<4; k++)
                        bz[tmp[k].x][tmp[k].y]=1;

                    int tx=tmp[j].x+dx[i];
                    int ty=tmp[j].y+dy[i];
                    if (!bound(tx,ty))  continue;
                    if (bz[tx][ty] )
                    {
                        tx=tx+dx[i];
                        ty=ty+dy[i];
                    }
                    if (!bound(tx,ty))  continue;
                    if (bz[tx][ty])  continue;
                    tmp[j].x=tx; tmp[j].y=ty;

                    data v;
                    v.s=change( tmp );
                    if ( s2.count(v.s) ) continue;
                    if (s1.count(v.s))
                    {
                        flag=1;
                        break;
                    }
                    s2.insert( v.s );
                    v.idx=u.idx+1;
                    if (v.idx<4)
                        q.push( v );
                }
                if (flag)  break;
            }
            if (flag)  break;
        }
        if (flag)  printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}