BZOJ 3170 [Tjoi2013]松鼠聚会

Posted 2020-10-24 ws_zzy

题解：切比雪夫距离转化为曼哈顿距离

枚举源点，横纵坐标互不影响，分开考虑，前缀和优化

横纵分开考虑是一种解题思路

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long Lint;
const int maxn=100009;

Lint ans=1000000000000000000LL;
int n;
int px[maxn],py[maxn];

int a[maxn];
Lint sa[maxn][2];

int b[maxn];
Lint sb[maxn][2];

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		int x,y;
		scanf("%d%d",&x,&y);
		px[i]=x+y;
		py[i]=x-y;
		a[i]=x+y;
		b[i]=x-y;
	}
	sort(a+1,a+1+n);
	sort(b+1,b+1+n);
	
	for(int i=1;i<=n;++i){
		sa[i][0]=sa[i-1][0]+a[i]-a[1];
		sb[i][0]=sb[i-1][0]+b[i]-b[1];
	}
	for(int i=n;i>=1;--i){
		sa[i][1]=sa[i+1][1]+a[n]-a[i];
		sb[i][1]=sb[i+1][1]+b[n]-b[i];
	}
	
	for(int i=1;i<=n;++i){
		int pa=lower_bound(a+1,a+1+n,px[i])-a;
		int pb=lower_bound(b+1,b+1+n,py[i])-b;
		Lint tm=0;
		tm+=(sa[1][1]-sa[pa][1])-1LL*(a[n]-a[pa])*(pa-1);
		tm+=(sa[n][0]-sa[pa][0])-1LL*(a[pa]-a[1])*(n-pa);
		tm+=(sb[1][1]-sb[pb][1])-1LL*(b[n]-b[pb])*(pb-1);
		tm+=(sb[n][0]-sb[pb][0])-1LL*(b[pb]-b[1])*(n-pb);
		ans=min(ans,tm);
	}
	
	printf("%lld\n",ans/2);
	return 0;
}