POJ1753 BFS+位运算

题意：翻转某些牌 使得达到目标状态

BFS+位运算

因为整个图比较小，所以用位运算来记录状态。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
using namespace std;
const int inf = 0x7fffffff;
const int maxn = 16;
const int maxm = 65535;
int vis[ maxm+5 ];
int dis[ maxm+5 ];
int mat[ maxn ][ maxn ];
int bina[]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768};
int cc[16]={19,39,78,140,305,626,1252,2248,4880,10016,20032,35968,12544,29184,58368,51200};
int main( int argc,char argv[] ){
    char a[ maxn ];
    int sum=0;
    for( int i=0;i<4;i++ ){
        scanf("%s",a);
        for( int j=0;j<4;j++ ){
            if( a[j]=='b' ){
                mat[i][j]=1;
            }
            else{
                mat[i][j]=0;
            }
            sum+=(bina[4*i+j]*mat[i][j]);
        }
    }
    memset( vis,0,sizeof(vis) );
    for( int i=0;i<=maxm;i++ )
        dis[ i ]=inf;
    if( sum==0||sum==maxm ){
        vis[ sum ]=1;
        dis[ sum ]=0;
    }
    queue<int>q;
    vis[ sum ]=1;
    dis[ sum ]=0;
    q.push( sum );
    while( !q.empty() ){
        int now=q.front();
        q.pop();
        if( now==0||now==maxm ){
            vis[ now ]=1;
            break;
        }
        if( vis[0]||vis[ maxm ] ){
            break;
        }
        for( int i=0;i<16;i++ ){
            int next=now^cc[i];
            if( vis[ next ]==0 ){
                vis[ next ]=1;
                dis[ next ]=dis[ now ]+1;
                q.push( next );
            }
            else{
                if( dis[ next ]>dis[ now ]+1 ){
                    dis[ next ]=dis[ now ]+1;
                }
            }
        }
    }
    if( vis[ 0 ]==1 ||vis[ maxm ]==1 ){
        printf("%d\n",min( dis[0],dis[ maxm ] ) );
    }
    else
    {
          printf("Impossible\n");
    }
    return 0;
}