1455.Solitaire（bfs状态混摇）

Solitaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3570    Accepted Submission(s): 1098

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

 

#include<stdio.h>
#include<queue>
#include<string.h>
#include<math.h>
#include<algorithm>
typedef long long ll ;
bool vis[8][8][8][8][8][8][8][8] ;
int move[][2] = {{1,0} , {-1,0} , {0,1} , {0,-1}} ;
struct no
{
    int x , y ;
} ee[4] ;

bool cmp (const no a , const no b)
{
    if (a.x < b.x) {
        return true ;
    }
    else if (a.x == b.x) {
        if (a.y < b.y) return true ;
        else return false ;
    }
    else return false ;
}

struct node
{
    int a[4][2] ;
    int step ;
};
node st , end ;

bool bfs ()
{
    node ans , tmp ;
    no rr[4] ;
    std::queue <node> q ;
    while ( !q.empty ()) q.pop () ;
    st.step = 0 ;
    q.push (st) ;
    while ( !q.empty ()) {
        ans = q.front () ; q.pop () ;
        for (int i = 0 ; i < 4 ; i ++) {
            for (int j = 0 ; j < 4 ; j ++) {
                tmp = ans ;
                int x = tmp.a[i][0] + move[j][0] , y = tmp.a[i][1] + move[j][1] ;
                if (x < 0 || y < 0 || x >= 8 || y >= 8) continue ;
                bool flag = 0 ;
                for (int i = 0 ; i < 4 ; i ++ ) {
                    if (x == tmp.a[i][0] && y == tmp.a[i][1])
                        flag = 1 ;
                }
                if (flag ) {
                    x += move[j][0] , y += move[j][1] ;
                    if (x < 0 || y < 0 || x >= 8 || y >= 8) continue ;
                    for (int i = 0 ; i < 4 ; i ++) {
                        if (x == tmp.a[i][0] && y == tmp.a[i][1])
                            flag = 0 ;
                    }
                    if ( !flag ) continue ;
                }
                tmp.step ++ ;
                if (tmp.step > 8) continue ;
                tmp.a[i][0] = x , tmp.a[i][1] = y ;
                if ( vis[tmp.a[0][0]] [tmp.a[0][1]] [tmp.a[1][0]] [tmp.a[1][1]] [tmp.a[2][0]] [tmp.a[2][1]] [tmp.a[3][0]] [tmp.a[3][1]]) continue;
                vis[tmp.a[0][0]] [tmp.a[0][1]] [tmp.a[1][0]] [tmp.a[1][1]] [tmp.a[2][0]] [tmp.a[2][1]] [tmp.a[3][0]] [tmp.a[3][1]] = 1 ;
                int cnt = 0 ;
                for (int i = 0 ; i < 4 ; i ++) {
                    rr[i].x = tmp.a[i][0] ; rr[i].y = tmp.a[i][1] ;
                }
                std::sort (rr , rr + 4 , cmp ) ;
              /*  for (int i = 0 ; i < 4 ; i ++) {
                    printf ("(%d , %d) , " , rr[i].x , rr[i].y ) ;
                } puts ("") ; */
                for (int i = 0 ; i < 4 ; i ++) {
                    if (rr[i].x != ee[i].x || rr[i].y != ee[i].y )
                        cnt ++ ;
                }
                if ( ! cnt ) return true ;
                if (tmp.step + cnt > 9) continue ;
                q.push (tmp) ;
               // printf ("step = %d
" , tmp.step ) ;
            }
        }
    }
    return false ;
}

int main ()
{
    //freopen ("a.txt" , "r" , stdin ) ;
    while ( ~ scanf ("%d%d" , &st.a[0][0] , &st.a[0][1])) {
        st.a[0][0] -- ; st.a[0][1] -- ;
        memset (vis , 0 , sizeof(vis)) ;
        for (int i = 1 ; i < 4 ; i ++) for (int j = 0 ; j < 2 ; j ++) { scanf ("%d" , &st.a[i][j]) ; st.a[i][j] -- ;}
        for (int i = 0 ; i < 4 ; i ++) for (int j = 0 ; j < 2 ; j ++) { scanf ("%d" , &end.a[i][j]) ; end.a[i][j] -- ;}
        for (int i = 0 ; i < 4 ; i ++) {
            ee[i].x = end.a[i][0] , ee[i].y = end.a[i][1] ;
        }
    // std::sort (ss , ss + 4 , cmp ) ;
        std::sort (ee , ee + 4 , cmp ) ;
        if ( bfs ()) puts ("YES") ;
        else puts ("NO") ;
    }
    return 0 ;
}

四枚棋子彼此不可分，所以对每次个状态因进行一下操作，比如说排序。