3170: [Tjoi 2013]松鼠聚会 (中位数)

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对于这题，我们可以得到d(i,j)=max(|xi-xj|,|yi-yj|)，

设x'=(x+y)/2,y'=(x-y)/2

那么d(i,j)=|xi'-xj'|+|yi'-yj'|

x，y轴可以分开统计

先是统计x轴，将所有松鼠的x'排序

用前缀和和后缀和求出某个松鼠到其他松鼠的X轴距离

Xi*(i-1)-sum(1..i-1)+sum(i+1..n)-(n-i)*Xi

y轴同理

#include<iostream>#include<cstdio>#include<algorithm>#define N 100005#define ll long longusing namespace std;int n;struct point{ll x,y;int num;}a[N];double ans=1e20;ll sx,sy,ax[N],bx[N],ay[N],by[N];bool cmpx(point a,point b){return a.x<b.x;}bool cmpy(point a,point b){return a.y<b.y;}int main(){scanf("%d",&n);for(int i=1;i<=n;i++){   ll x,y;   scanf("%lld%lld",&x,&y);   a[i].x=x+y;a[i].y=x-y;    }sort(a+1,a+n+1,cmpy);for(int i=1;i<=n;i++){ay[i]=ay[i-1]+a[i].y;by[n-i+1]=by[n-i+2]-a[n-i+1].y;a[i].num=i;}sort(a+1,a+n+1,cmpx);for(int i=1;i<=n;i++){ax[i]=ax[i-1]+a[i].x;bx[n-i+1]=bx[n-i+2]-a[n-i+1].x;}for(int i=1;i<=n;i++)    {        ll tmp=0;        tmp+=ax[n]-ax[i]-(n-i)*(ax[i]-ax[i-1]);        tmp+=bx[1]-bx[i]-(i-1)*(bx[i]-bx[i+1]);        int j=a[i].num;        tmp+=ay[n]-ay[j]-(n-j)*(ay[j]-ay[j-1]);        tmp+=by[1]-by[j]-(j-1)*(by[j]-by[j+1]);        if(tmp<ans)ans=tmp;    }    printf("%.0lf",ans/2);return 0;}

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