BZOJ2674 : Attack

整体二分+树状数组套Treap,时间复杂度$O(n\log^3n)$。

 

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define N 100010
using namespace std;
int n,m,tot,i,X0,Y0,X1,Y1,cq,bx[N],by[N],bz[N],ans[N],tmp,C,D;char op[8];
struct P{
  int a,b,c,d,k,p;
  P(){}
  P(int _a,int _b,int _c,int _d,int _k,int _p){a=_a,b=_b,c=_c,d=_d,k=_k,p=_p;}
}a[N],b[N],ql[N],qr[N];
inline int findl(int*a,int x){
  int l=1,r=n,mid,t;
  while(l<=r)if(a[mid=(l+r)>>1]>=x)r=(t=mid)-1;else l=mid+1;
  return t;
}
inline int findr(int*a,int x){
  int l=1,r=n,mid,t;
  while(l<=r)if(a[mid=(l+r)>>1]<=x)l=(t=mid)+1;else r=mid-1;
  return t;
}
struct node{
  int p,val,v,sum;node*l,*r;
  node(){val=v=sum=p=0;l=r=NULL;}
  inline void up(){sum=v+l->sum+r->sum;}
}*blank=new(node),*bit[N],pool[6000000],*cur=pool;int pos[N],T;
inline void Rotatel(node*&x){node*y=x->r;x->r=y->l;x->up();y->l=x;y->up();x=y;}
inline void Rotater(node*&x){node*y=x->l;x->l=y->r;x->up();y->r=x;y->up();x=y;}
void Ins(node*&x){
  if(x==blank){
    x=cur++;x->val=C;x->l=x->r=blank;x->v=x->sum=D;x->p=std::rand();
    return;
  }
  x->sum+=D;
  if(C==x->val){x->v+=D;return;}
  if(C<x->val){
    Ins(x->l);
    if(x->l->p>x->p)Rotater(x);
  }else{
    Ins(x->r);
    if(x->r->p>x->p)Rotatel(x);
  }
}
void Ask(node*&x,int a,int b){
  if(x==blank)return;
  if(C<=a&&b<=D){tmp+=x->sum;return;}
  if(C<=x->val&&x->val<=D)tmp+=x->v;
  if(C<x->val)Ask(x->l,a,x->val-1);
  if(D>x->val)Ask(x->r,x->val+1,b);
}
inline void add(int x,int y,int p){for(C=y,D=p;x<=n;Ins(bit[x]),x+=x&-x)if(pos[x]<T)pos[x]=T,bit[x]=blank;}
inline int sum(int x,int l,int r){for(C=l,D=r,tmp=0;x;x-=x&-x)if(pos[x]==T)Ask(bit[x],0,n);return tmp;}
void solve(int l,int r,int L,int R){
  if(L>R)return;
  if(l==r){
    for(int i=L;i<=R;i++)if(b[i].p)ans[b[i].p]=l;
    return;
  }
  int mid=(l+r)>>1,i,cl=0,cr=0;
  for(T++,i=L;i<=R;i++)if(!b[i].p){
    if(b[i].k<=mid)add(b[i].a,b[i].b,b[i].c),ql[cl++]=b[i];else qr[cr++]=b[i];
  }else{
    int t=sum(b[i].b,b[i].c,b[i].d)-sum(b[i].a-1,b[i].c,b[i].d);
    if(t>=b[i].k)ql[cl++]=b[i];else b[i].k-=t,qr[cr++]=b[i];
  }
  for(i=0;i<cl;i++)b[L+i]=ql[i];
  for(i=0;i<cr;i++)b[L+cl+i]=qr[i];
  solve(l,mid,L,L+cl-1),solve(mid+1,r,L+cl,R);
}
int main(){
  scanf("%d%d",&n,&m);
  for(i=1;i<=n;i++){
    scanf("%d%d%d",&a[i].a,&a[i].b,&a[i].k);
    bx[i]=a[i].a,by[i]=a[i].b,bz[i]=a[i].k;
  }
  sort(bx+1,bx+n+1),sort(by+1,by+n+1),sort(bz+1,bz+n+1);
  for(i=1;i<=n;i++){
    a[i].a=findl(bx,a[i].a);
    a[i].b=findl(by,a[i].b);
    a[i].k=findl(bz,a[i].k);
    a[i].c=1;
    b[i]=a[i];
  }
  tot=n;
  while(m--){
    scanf("%s%d%d",op,&X0,&Y0);
    if(op[0]=='S'){
      X0++,Y0++;
      if(X0==Y0)continue;
      b[++tot]=P(a[X0].a,a[X0].b,-1,0,a[X0].k,0);
      b[++tot]=P(a[X0].a,a[X0].b,1,0,a[Y0].k,0);
      b[++tot]=P(a[Y0].a,a[Y0].b,1,0,a[X0].k,0);
      b[++tot]=P(a[Y0].a,a[Y0].b,-1,0,a[Y0].k,0);
      swap(a[X0].k,a[Y0].k);
    }else{
      scanf("%d%d%d",&X1,&Y1,&i);
      if(X0>X1)swap(X0,X1);
      if(Y0>Y1)swap(Y0,Y1);
      if(X0>bx[n]||X1<bx[1]||Y0>by[n]||Y1<by[1])X0=1,X1=Y0=Y1=0;
      else{
        X0=findl(bx,X0);
        X1=findr(bx,X1);
        Y0=findl(by,Y0);
        Y1=findr(by,Y1);
      }
      b[++tot]=P(X0,X1,Y0,Y1,i,++cq);
    }
  }
  solve(1,n+1,1,tot);
  for(i=1;i<=cq;i++)if(ans[i]>n)puts("It doesn't exist.");else printf("%d\n",bz[ans[i]]);
  return 0;
}

  

posted @ 2015-10-11 22:51  Claris  阅读(368)  评论(0编辑  收藏  举报