洛咕 P3964 [TJOI2013]松鼠聚会

有个结论就是把坐标\((x,y)\)变形成\(((x+y)/2,(x-y)/2)\)，切比雪夫距离就变成了曼哈顿距离。

所以变换一下坐标直接统计答案即可。

// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define il inline
#define vd void
typedef long long ll;
il int gi(){
    int x=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
struct yyb{ll x,y;}s[100010];
il bool operator <(const yyb&a,const yyb&b){return a.x<b.x;}
ll Y[100010],sY[100010],sX[100010];
int main(){
#ifndef ONLINE_JUDGE
    freopen("3964.in","r",stdin);
    freopen("3964.out","w",stdout);
#endif
    int n=gi();
    for(int i=1;i<=n;++i){
        int x=gi(),y=gi();
        s[i].x=x+y,s[i].y=x-y;
        Y[i]=s[i].y;
    }
    std::sort(s+1,s+n+1);
    std::sort(Y+1,Y+n+1);
    for(int i=1;i<=n;++i)sY[i]=sY[i-1]+Y[i];
    for(int i=1;i<=n;++i)sX[i]=sX[i-1]+s[i].x;
    ll ans=1e18;
    for(int i=1;i<=n;++i){
        int p=std::lower_bound(Y+1,Y+n+1,s[i].y)-Y;
        ans=std::min(ans,-sX[i]*2+(i+i-n)*s[i].x+sX[n]-sY[p]*2+(p+p-n)*s[i].y+sY[n]);
    }
    printf("%lld\n",ans/2);
    return 0;
}