[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(n^2)。

 #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=,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 ;
     if (k2<i || k2>i+n) return ;
     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)>>,pos=;
     rep(i,L,R) if (chk(mid,i,pos)) pos=i;
     ans[mid]=pos>n ? pos-n : pos;
     solve(l,mid-,L,pos); solve(mid+,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,,n) scanf("%d%d",&a[i].x,&a[i].y),a[i].id=i,a[n+i]=a[i];
         solve(,n,,n<<);
         rep(i,,n) printf("%d\n",ans[i]);
     }
     return ;
 }