BZOJ3170: [Tjoi 2013]松鼠聚会

切比雪夫距离，直接上公式转成哈密顿距离
然后就是找一个点到其他所有点的哈密顿距离和最短，将两维分开考虑，排个序O(n)就可以算出每个点到其他所有点x坐标的距离和，y坐标的距离和

注意longlong…

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#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long longusing namespace std;const ll maxn = 110000;struct node{    ll x,i;}p1[maxn],p2[maxn];ll s1[maxn],s2[maxn];bool cmp(node x,node y){return x.x<y.x;}ll n;int main(){    scanf("%lld",&n);    for(ll i=1;i<=n;i++)    {        ll x,y; scanf("%lld%lld",&x,&y);        p1[i].x=x+y; p1[i].i=i;         p2[i].x=x-y; p2[i].i=i;    }    sort(p1+1,p1+n+1,cmp);    sort(p2+1,p2+n+1,cmp);    for(ll i=2;i<=n;i++)    {        s1[p1[1].i]+=abs(p1[1].x-p1[i].x);        s2[p2[1].i]+=abs(p2[1].x-p2[i].x);    }    for(ll i=2;i<=n;i++)    {        s1[p1[i].i]=s1[p1[i-1].i]-(p1[i].x-p1[i-1].x)*(n-i+1-i+1);        s2[p2[i].i]=s2[p2[i-1].i]-(p2[i].x-p2[i-1].x)*(n-i+1-i+1);    }    ll ans=1;    for(ll i=2;i<=n;i++)    {        if(s1[i]+s2[i]<s1[ans]+s2[ans]) ans=i;    }    printf("%lld\n",(s1[ans]+s2[ans])>>1);    return 0;}