topic
analyze
We define \(dis_{x,y,x1,y2}\) to represent the distance from \((x,y)\) to \((x1,y1)\) . Use spfa for this.
Next, enumerate the two rendezvous points \((x,y), (x1,y1)\) , get the distance from these two rendezvous points to the knight, put \(dis1\) and \(dis2\) .
Then consider greed,
assuming that all \(dis1\) are selected , and the sum is \(sum\) . Sort \(dis2- dis1 \) from small to large, add the value of the previous \(\dfrac{n}{2}\) to \(sum\) , the answer is \(\min(sum)\)
#include <cmath>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
const int maxlongint=2147483647;
const int mo=1000000007;
const int N=205;
using namespace std;
int dis[21][21][21][21],a[N],n,m,r,c,ans=maxlongint,b[N][2],d[N*100][4],tot;
int z[8][2]=
{
{2,1},
{2,-1},
{-2,1},
{-2,-1},
{1,2},
{1,-2},
{-1,-2},
{-1,2}
};
bool bz[21][21];
int spfa(int x,int y)
{
int head=0,tail=1,xx,yy;
d[1][1]=x;
d[1][2]=y;
dis[x][y][x][y]=0;
while(head<tail)
{
xx=d[++head][1];
yy=d[head][2];
bz[xx][yy]=true;
for(int i=0;i<=7;i++)
{
if(dis[x][y][xx+z[i][0]][yy+z[i][1]]>dis[x][y][xx][yy]+1)
{
dis[x][y][xx+z[i][0]][yy+z[i][1]]=dis[x][y][xx][yy]+1;
if(bz[xx+z[i][0]][yy+z[i][1]])
{
bz[xx+z[i][0]][yy+z[i][1]]=false;
d[++tail][1]=xx+z[i][0];
d[tail][2]=yy+z[i][1];
}
}
}
}
}
int main()
{
scanf("%d%d%d",&n,&r,&c);
for(int i=1;i<=n;i++)
for(int j=0;j<=1;j++)
scanf("%d",&b[i][j]);
memset(dis,43,sizeof(dis));
for(int i=1;i<=r;i++)
for(int j=1;j<=c;j++)
{
memset(bz,true,sizeof(bz));
spfa(i,j);
}
for(int x=1;x<=r;x++)
for(int y=1;y<=c;y++)
{
int sum=0;
for(int i=1;i<=n;i++)
{
sum+=dis[x][y][b[i][0]][b[i][1]];
}
int o=sum;
for(int xx=1;xx<=r;xx++)
for(int yy=1;yy<=c;yy++)
if(x!=xx || y!=yy)
{
sum=o;
tot=0;
for(int i=1;i<=n;i++)
{
a[++tot]=dis[xx][yy][b[i][0]][b[i][1]]-dis[x][y][b[i][0]][b[i][1]];
}
sort(a+1,a+tot+1);
for(int i=1;i<=n/2;i++)
{
sum+=a[i];
}
if(sum<ans)
ans=sum;
}
}
printf("%d",ans);
}