这题的意思是找到最少的时间和使所有Knight和King到达一个格子...但蛋疼的是King不仅可以按平时的规则走...在和Knight相遇后可以和骑士合体一起走...=.=.....
先做好两两点之间通过Knight的步伐能走到的最短时间这里可以打出一个表..4维的..arc[x1][y1][x2][y2] 代表(x1,y1)到(x2,y2)的"Knight"最短距离....再开始枚举每个点..枚举每个点时又枚举是王自己走过去还是王走几步后和哪个骑士一起过去...好吧..其实我的程序也是大概来猜测的..给了King最多4步的移动空间~~但似乎还是可以过所有数据的...
Program:
/*
ID: zzyzzy12
LANG: C++
TASK: camelot
*/
#include<iostream>
#include<istream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stack>
#include<algorithm>
#include<queue>
#define oo 1000000000
#define ll long long
using namespace std;
struct node
{
int x,y,w;
}h,k,king,knight[1000];
queue<node> myqueue;
int R,C,p,arc[33][33][33][33],kingm[33][33],num;
int move[2][8][2]={{{1,0},{1,1},{0,1},{-1,1},{-1,0},{1,-1},{0,-1},{-1,-1}},
{{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1},{-2,1},{-1,2}}};
int getdata()
{
char c='1';
while (c<'A' || c>'Z' )
if (scanf("%c",&c)==EOF) return 0;
return c-'A'+1;
}
void make(node hh,int tp)
{
h=hh;
arc[hh.x][hh.y][h.x][h.y]=0;
myqueue.push(h);
while (!myqueue.empty())
{
h=myqueue.front();
myqueue.pop();
for (p=0;p<8;p++)
{
k.x=h.x+move[tp][p][0];
k.y=h.y+move[tp][p][1];
if (k.x<1 || k.y<1 || k.x>C || k.y>R) continue;
if (arc[hh.x][hh.y][k.x][k.y]==-1)
{
k.w=h.w+1;
myqueue.push(k);
arc[hh.x][hh.y][k.x][k.y]=k.w;
}
}
}
}
void makekingmove()
{
h=king;
kingm[h.x][h.y]=0;
myqueue.push(h);
while (!myqueue.empty())
{
h=myqueue.front();
myqueue.pop();
for (p=0;p<8;p++)
{
k.x=h.x+move[0][p][0];
k.y=h.y+move[0][p][1];
if (k.x<1 || k.y<1 || k.x>C || k.y>R) continue;
if (kingm[k.x][k.y]==-1)
{
k.w=h.w+1;
myqueue.push(k);
kingm[k.x][k.y]=k.w;
}
}
}
}
int getanswer()
{
int x,y,p,i,j,m,k,t,v,ans=oo;
for (x=1;x<=C;x++)
for (y=1;y<=R;y++)
{
m=kingm[x][y];
for (i=1;i<=num;i++)
if (arc[knight[i].x][knight[i].y][x][y]==-1) goto A;
else m+=arc[knight[i].x][knight[i].y][x][y];
if (m<ans) ans=m;
for (t=0;t<5;t++)
for (v=0;v<8;v++)
{
i=move[0][v][0]*t+king.x; j=move[0][v][1]*t+king.y;
if (i<1 || j<1 || i>C || j>R) continue;
if (arc[i][j][x][y]!=-1)
{
for (p=1;p<=num;p++)
if (arc[knight[p].x][knight[p].y][i][j]!=-1)
{
k=m-kingm[x][y]-arc[knight[p].x][knight[p].y][x][y]+arc[knight[p].x][knight[p].y][i][j]+arc[i][j][x][y]+kingm[i][j];
if (k<ans) ans=k;
}
}
}
A: ;
}
return ans;
}
int main()
{
freopen("camelot.in","r",stdin);
freopen("camelot.out","w",stdout);
while (!myqueue.empty()) myqueue.pop();
memset(arc,-1,sizeof(arc));
memset(kingm,-1,sizeof(kingm));
scanf("%d%d",&R,&C);
king.x=getdata(); scanf("%d",&king.y); king.w=0;
makekingmove();
for (int x=1;x<=C;x++)
for (int y=1;y<=R;y++)
{
h.x=x; h.y=y; h.w=0;
make(h,1);
}
num=0;
while (1)
{
h.x=getdata();
if (!h.x) break;
scanf("%d",&h.y); h.w=0;
knight[++num]=h;
}
printf("%d\n",getanswer());
return 0;
}