[BZOJ2739]最远点(DP+分治+决策单调性)

根据旋转卡壳，当逆时针遍历点时，相应的最远点也逆时针转动，满足决策单调性。于是倍长成链，分治优化DP即可，复杂度O(nlogn)。

#include<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;

const int N=1000010,inf=1e9;

int n,T,ans[N];
struct P{ int x,y,id; }a[N];
ll sqr(ll x){ return x*x; }
ll dis(P a,P b){ return sqr(a.x-b.x)+sqr(a.y-b.y); }

bool chk(int i,int k1,int k2){
    if (k1<i || k1>i+n) return 0;
    if (k2<i || k2>i+n) return 1;
    ll d1=dis(a[i],a[k1]),d2=dis(a[i],a[k2]);
    return d1==d2 ? a[k1].id<a[k2].id : d1>d2;
}

void solve(int l,int r,int L,int R){
    if (l>r) return;
    int mid=(l+r)>>1,pos=0;
    rep(i,L,R) if (chk(mid,i,pos)) pos=i;
    ans[mid]=pos>n ? pos-n : pos;
    solve(l,mid-1,L,pos); solve(mid+1,r,pos,R);
}

int main(){
    freopen("bzoj2739.in","r",stdin);
    freopen("bzoj2739.out","w",stdout);
    for (scanf("%d",&T); T--; ){
        scanf("%d",&n);
        rep(i,1,n) scanf("%d%d",&a[i].x,&a[i].y),a[i].id=i,a[n+i]=a[i];
        solve(1,n,1,n<<1);
        rep(i,1,n) printf("%d\n",ans[i]);
    }
    return 0;
}