BZOJ2149 : 拆迁队

设$c[i]=g[i]+\frac{i(i+1)}{2}-a[i]\times i-a[i]$，$d[i]=a[i]-i$

$f[i]$表示以$i$为结尾最多保留多少个建筑，则

$f[i]=\max(f[j])+1$，$j<i且d[j]\leq d[i]$

$g[i]$表示以$i$为结尾的最小代价，则

$g[i]=\min(i\times d[j]+c[j])+b[i]+a[i]+\frac{i(i-1)}{2}$，$j<i,f[j]+1=f[i]且d[j]\leq d[i]$

按$f$一组一组来处理，每组按序号排序，然后分治处理。

分治的时候，按$d$排序，用栈维护凸壳，查询时在凸壳上二分。

时间复杂度$O(n\log^2n)$。

#include<cstdio>
#include<algorithm>
#define N 100010
using namespace std;
typedef long long ll;
int n,m,i,j,k,a[N],b[N],d[N],e[N],bit[N],f[N],ans0,G[N],nxt[N];
int cl,cr,L[N],R[N],top,q[N];
ll c[N],g[N],ans1=1LL<<62;
struct E{int x,t;E(){}E(int _x,int _t){x=_x,t=_t;}}h[N];
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
inline void umax(int&a,int b){if(a<b)a=b;}
inline void umin(ll&a,ll b){if(a>b)a=b;}
inline int lower(int x){
  int l=1,r=n+1,mid,t;
  while(l<=r)if(e[mid=(l+r)>>1]<=x)l=(t=mid)+1;else r=mid-1;
  return t;
}
inline void ins(int x,int y){for(;x<=n+1;x+=x&-x)umax(bit[x],y);}
inline int ask(int x){int t=-N;for(;x;x-=x&-x)umax(t,bit[x]);return t;}
inline void add(int x,int y){nxt[y]=G[x];G[x]=y;}
inline bool cmp(int x,int y){return d[x]==d[y]?c[x]>c[y]:d[x]<d[y];}
inline double pos(int x,int y){return 1.0*(c[y]-c[x])/(d[x]-d[y]);}
inline void insert(int x){
  if(top&&d[x]==d[q[top]])top--;
  while(top>1&&pos(q[top],x)>pos(q[top],q[top-1]))top--;
  q[++top]=x;
}
inline void query(int x){
  if(!top)return;
  int l=1,r=top-1,mid,t=top;
  while(l<=r){
    mid=(l+r)>>1;
    if(x>pos(q[mid],q[mid+1]))r=(t=mid)-1;else l=mid+1;
  }
  umin(g[x],1LL*x*d[q[t]]+c[q[t]]);
}
void solve(int l,int r){
  if(l==r)return;
  int mid=(l+r)>>1,i,j;
  solve(l,mid),solve(mid+1,r);
  for(cl=0,i=l;i<=mid;i++)if(!h[i].t)L[++cl]=h[i].x;
  for(cr=0,i=r;i>mid;i--)if(h[i].t)R[++cr]=h[i].x;
  if(!cl||!cr)return;
  sort(L+1,L+cl+1,cmp),sort(R+1,R+cr+1,cmp);
  for(i=j=1,top=0;i<=cr;i++){
    while(j<=cl&&d[L[j]]<=d[R[i]])insert(L[j++]);
    query(R[i]);
  }
}
int main(){
  read(n);
  for(i=1;i<=n;i++)read(a[i]);
  for(i=1;i<=n;i++)read(b[i]),e[i+1]=d[i]=a[i]-i;
  sort(e+1,e+n+2);
  for(i=1;i<=n+1;i++)bit[i]=-N;
  ins(lower(0),0);
  for(i=1;i<=n;i++)g[i]=ans1,j=lower(d[i]),ins(j,f[i]=ask(j)+1);
  ans0=ask(n+1);
  for(i=0;i<=n;i++)G[i]=-1;
  for(i=n;~i;i--)if(f[i]>=0)add(f[i],i);
  for(i=0;i<ans0;i++){
    j=G[i],k=G[i+1],m=0;
    while(~j||~k){
      if(j<0)h[++m]=E(k,1),k=nxt[k];
      else if(k<0||j<k)h[++m]=E(j,0),j=nxt[j];
      else h[++m]=E(k,1),k=nxt[k];
    }
    solve(1,m);
    for(k=G[i+1];~k;k=nxt[k]){
      g[k]+=b[k]+a[k]+1LL*k*(k-1)/2;
      c[k]=g[k]+1LL*k*(k+1)/2-1LL*a[k]*k-a[k];
    }
  }
  for(i=0;i<=n;i++)if(f[i]==ans0)umin(ans1,g[i]+1LL*(n-i)*a[i]+1LL*(n-i)*(n-i+1)/2);
  return printf("%d %lld",ans0,ans1),0;
}