学步园

骑士走的方式用普通的BFS，保存每个骑士的最短路径。然后枚举每个骑士在每个点接国王，计算路径即可。

相当麻烦的考验耐心的题目。

/*
ID:bysen
LANG:C++
PROG:camelot
*/
#include<stdio.h>
#include<queue>
#define MAXR 31
#define MAXC 27
#define INF 0x0FFFFFFF
using namespace std;

struct K
{
 	   int r,c;
}knight[MAXR*MAXC];

K king;


struct bfsK
{
 	   int r,c,step;
};

int knightnum;
int remap[MAXR*MAXR][MAXR][MAXC];
int R,C;

int max( int a,int b ){ return a>b?a:b; }

//坐标映射 
int getR( int num ){  return num/C+1; }
int getC( int num ){  return num%C+1; }
int getNum( int r,int c ){ return (r-1)*C+c-1; }
int d[8][2]={ {-2,+1},{-1,+2},{+1,+2},{+2,+1},{+2,-1},{+1,-2},{-1,-2},{-2,-1} };

void init()
{
 	freopen( "camelot.in","r",stdin );
 	freopen( "camelot.out","w",stdout );
 	scanf( "%d %d\n",&R,&C );
 	char c;int r;
 	
 	for( int k=0;k<MAXR*MAXC;k++ )
 	for( int i=0;i<=R;i++ )
 	for( int j=0;j<=C;j++ )
 		 remap[k][i][j]=INF;
 	
 	scanf( "%c %d",&c,&r );
 	king.r=r;king.c=c-'A'+1;
 	
 	knightnum=0;
 	
 	while( scanf( "%c",&c )!=EOF )
 	{
	 	   if( c<'A' || c>'Z' )
	 	   	   continue;
	 	   scanf( "%d\n",&r );
		   knight[knightnum].r=r;
		   knight[knightnum].c=c-'A'+1;
		   knightnum++;
    }
    /*
    for( int i=0;i<knightnum;i++ )
    	 printf( "%d: %d %d\n",i,knight[i].r,knight[i].c );
    	 */
}

void bfs( int num )
{
 	 int r,c,x,y;
 	 bool flag[MAXR][MAXC]={false};
 	 r=getR(num);
 	 c=getC(num);
 	 bfsK k;
 	 k.r=r;k.c=c;k.step=0;
 	 queue<bfsK>myQueue;
 	 myQueue.push(k);
 	 flag[k.r][k.c]=true;
 	 remap[num][k.r][k.c]=0;
 	 
	 while( !myQueue.empty() )
	 {
	  		bfsK cur=myQueue.front();
	  		myQueue.pop();
	  		r=cur.r;c=cur.c;
	  		for( int i=0;i<8;i++ )
	  		{
			 	 x=d[i][0];y=d[i][1];
			 	 if( r+x>=1 && r+x<=R && c+y>=1 && c+y<=C && flag[r+x][c+y]==false )
			 	 {
				  	 flag[r+x][c+y]=true;
				  	 bfsK tmp;
				  	 tmp.r=r+x;
				  	 tmp.c=c+y;
				  	 tmp.step=cur.step+1;
				  	 remap[num][tmp.r][tmp.c]=tmp.step;
				  	 myQueue.push(tmp);
  	 		     }
 	 	    }
     }
}

void record( )
{
 	 for( int k=0;k<R*C;k++ )
	  	  bfs( k );
	  /*
	 for( int k=0;k<R*C;k++ )
	 {
	  	  printf( "%d\n",k );
	  	  for( int i=1;i<=R;i++ )
	  	  {
	  	   	   for( int j=1;j<=C;j++ )
	  	  	   		printf( "%d ",remap[k][i][j] );
	  	  	   printf( "\n" );
          }
     }
	 for( int k=0;k<knightnum;k++ )
	 {
	  	  printf( "%d %d:\n",knight[k].r,knight[k].c );
	  	  printf( "%d\n",getNum(knight[k].r,knight[k].c) );
	  	  for( int i=1;i<=R;i++ )
	  	  {
	  	  	   for( int j=1;j<=C;j++ )
	  	  	   		printf( "%d ",remap[getNum(knight[k].r,knight[k].c)][i][j] );
 	  	   	   printf( "\n" );
		  }
     }
	 */
}

void enumwork()
{
 	 int ans=INF;
 	 int sum[MAXR][MAXC]={0};
 	 
 	 for( int k=0;k<knightnum;k++ )
 	 {
	  	  int remapnum=getNum(knight[k].r,knight[k].c);
 	  	  for( int i=1;i<=R;i++ )
 	 	  for( int j=1;j<=C;j++ )
  	   	  	   sum[i][j]+=remap[remapnum][i][j];
	 }
	 /*
	 for( int i=1;i<=R;i++ )
	 {
	  	  for( int j=1;j<=C;j++ )
	 	  	   printf( "%d ",sum[i][j]);
  	      printf( "\n" );
     }
     */
 	 for( int i=1;i<=R;i++ )
 	 for( int j=1;j<=C;j++ )
 	 {
	  	  //枚举最终集合点 
	  	  int enumnum=getNum(i,j);
	  	  if( ans>sum[i][j] )
	  	  {
		   	  int knightAllStep=sum[i][j];
		   	  int kingstep;
				 //枚举 哪个骑士来接国王 
		   	  for( int k=0;k<knightnum;k++ )
		   	  {
			   	   int NoKnight=getNum(knight[k].r,knight[k].c);//骑士在预存图中的编号
				   int knightRideStep;//接国王的骑士走的步数 
			   	   knightAllStep=sum[i][j]-remap[NoKnight][i][j];
			   	   //除了这个骑士外其他骑士走的步数
			   	   //枚举骑士国王的碰头地点
				   for( int a=1;a<=R;a++ )
				   for( int b=1;b<=C;b++ )
				   {
				   		kingstep=max( abs(king.r-a),abs(king.c-b) );
				   		knightRideStep=remap[NoKnight][a][b]+remap[getNum(a,b)][i][j];//骑士从源点走向与王汇合点+从汇合点到所有人的集合点;
				   		if( ans>kingstep+knightRideStep+knightAllStep )
				   			ans=kingstep+knightRideStep+knightAllStep;
		   		   }
	   	      }
		  }
	 }
	 printf( "%d\n",ans );
}

int main()
{
 	init();
 	if( knightnum==0 )
 	{
 		printf( "0\n" );
 		return 0;
    }
 	record();//对每个位置进行一次
	enumwork();//枚举工作 
 	return 0;
}