洛谷P3964 [TJOI2013]松鼠聚会 切比雪夫距离转曼哈顿距离

https://www.luogu.com.cn/problem/P3964

切比雪夫距离为max(|x-o.x|,|y-o.y|)，曼哈顿距离为|x-o.x|+|y-o.y|

当我们要求n个点的到1个点的切比雪夫距离和最小，显然有个max值是不好求的，那么可以通过(x,y)->((x+y)/2,(x-y)/2)的坐标变换变成曼哈顿距离，然后分别求每个点为终点的x和y的距离之和，sort一下扫一遍就好了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxl=2e5+10;

int n;ll ans;
ll sumx[maxl],sumy[maxl];
struct node
{
	ll x,y;int id;
}a[maxl],b[maxl];

inline void prework()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld%lld",&a[i].x,&a[i].y);
		a[i].x*=2;a[i].y*=2;
		b[i].x=(a[i].x+a[i].y)/2;
		b[i].y=(a[i].x-a[i].y)/2;
		b[i].id=i;
	}
}

inline bool cmpx(const node &a,const node &b)
{
	return a.x<b.x;
}

inline bool cmpy(const node &a,const node &b)
{
	return a.y<b.y;
}

inline void mainwork()
{
	ll sum=0,cnt=n-1;
	sort(b+1,b+1+n,cmpx);
	for(int i=2;i<=n;i++)
		sum+=b[i].x-b[1].x;
	sumx[b[1].id]=sum;
	for(int i=2;i<=n;i++)
	{
		sum-=(n-i+1)*(b[i].x-b[i-1].x);
		sum+=(i-1)*(b[i].x-b[i-1].x);
		sumx[b[i].id]=sum;
	}
	sort(b+1,b+1+n,cmpy);
	sum=0;
	for(int i=2;i<=n;i++)
		sum+=b[i].y-b[1].y;
	sumy[b[1].id]=sum;
	for(int i=2;i<=n;i++)
	{
		sum-=(n-i+1)*(b[i].y-b[i-1].y);
		sum+=(i-1)*(b[i].y-b[i-1].y);
		sumy[b[i].id]=sum;
	}
	ans=sumx[1]+sumy[1];
	for(int i=2;i<=n;i++)
		ans=min(ans,sumx[i]+sumy[i]);
}

inline void print()
{
	printf("%lld\n",ans/2);
}

int main()
{
	prework();
	mainwork();
	print();
	return 0;
}