pku2965 The Pilots Brothers' refrigerator

 

第一道双向广搜,188MS

题意:给定一个4X4的“+”“-”图，定义操作[i,j]为改变第i行和第j列的状态，输出使原图变为全“-”图的最小的操作数和各具体操作

#include  < iostream >
using   namespace  std;

#define  MAXN 2<<17

int  r[ 5 ] = { 0 , 61440 , 3840 , 240 , 15 };
int  c[ 5 ] = { 0 , 34952 , 17476 , 8738 , 4369 };
int  rc[ 5 ][ 5 ] = {
    { 0 , 0 , 0 , 0 , 0 },
    { 0 , 32768 , 16384 , 8192 , 4096 },
    { 0 , 2048 , 1024 , 512 , 256 },
    { 0 , 128 , 64 , 32 , 16 },
    { 0 , 8 , 4 , 2 , 1 }
};

int  open[MAXN],head,tail,mk[MAXN],st,ed,opera[MAXN],pre[MAXN],ans[MAXN],cnt;

char  ss[ 6 ][ 6 ];

int  change( int  i, int  j, int  s){
     int  ns;
    
    ns =~ (r[i] ^ ( ~ s));
    ns =~ (c[j] ^ ( ~ ns));
    ns =~ (rc[i][j] ^ ( ~ ns));

     return  ns;
}

void  output( int  s, int  ns){
     if (mk[s] == 2 ){
         int  t = s;
        s = ns;
        ns = t;
    }
     while (s != st){
        ans[cnt ++ ] = opera[s];
        s = pre[s];
    };

     while (ns != ed){
        ans[cnt ++ ] = opera[ns];
        ns = pre[ns];
    };
}



void  solve(){
     int  i,j,s,ns;
     if (st == ed){
        printf( " 0\n " );
         return ;
    }
    open[ 0 ] = st;
    open[ 1 ] = ed;
    mk[st] = 1 ;
    mk[ed] = 2 ;
    head = 0 ;
    tail = 2 ;
     while (head < tail){

        s = open[head ++ ];
         for (i = 1 ;i <= 4 ;i ++ ){
             for (j = 1 ;j <= 4 ;j ++ ){
                ns = change(i,j,s);
                 if ( ! mk[ns]){
                    mk[ns] = mk[s];
                    opera[ns] = i * 10 + j;
                    pre[ns] = s;
                    open[tail ++ ] = ns;
                }
                 else {
                     if (mk[s] != mk[ns]){
                         // cout<<"!!\n";
                         // printf("%d %d\n",i,j);
                        ans[cnt ++ ] = i * 10 + j;
                        output(s,ns);
                         return ;
                    }
                }
            }
        }
    }
     return ;
}



int  main(){
    cnt = 0 ;
     int  i,j,k;
     for (i = 0 ;i < 4 ;i ++ )
        gets(ss[i]);

    st = ed = 0 ;
    k = 0 ;

     // 建立st的位图
     for (i = 3 ;i >= 0 ;i -- ){
         for (j = 3 ;j >= 0 ;j -- ){
             if (ss[i][j] == ' + ' )
                st += 1 << k;
            k ++ ;
        }
    }

    solve();
    printf( " %d\n " ,cnt);
     for (i = 0 ;i < cnt;i ++ )
        printf( " %d %d\n " ,ans[i] / 10 ,ans[i] % 10 );

     return   0 ;
}