bzoj 4604: The kth maximum number

题意：

题解：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
using namespace std;
struct node{
int op,x,y,c;
}Q[200010],q[200010],q1[200010],q2[200010],tmp[200010];int cnt=0,qcnt=0;
const int inf=1e9;
int n,m,K[50010],ans[50010],num[50010],tr[500010],re[200010],tim=0,C[50010];
void change(int k,int c) {for(int i=k;i<=n;i+=(i&-i)) tr[i]+=c;}
int get(int k) {int ans=0;for(int i=k;i>=1;i-=(i&-i)) ans+=tr[i];return ans;}
int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
void cdq(int l,int r,int Mid)
{
if(l==r) return;
int mid=(l+r)/2;
cdq(l,mid,Mid);cdq(mid+1,r,Mid);
int i=l,j=mid+1,len=0,st=0;
while(i<=mid&&j<=r)
{
if(Q[i].x<=Q[j].x)
{
if(Q[i].op==1&&Q[i].c>Mid) change(Q[i].y,1),re[++st]=Q[i].y;
tmp[++len]=Q[i++];
}
else
{
if(Q[j].op==2) num[Q[j].c]+=get(Q[j].y);
if(Q[j].op==3) num[Q[j].c]-=get(Q[j].y);
tmp[++len]=Q[j++];
}
}
while(i<=mid) tmp[++len]=Q[i++];
while(j<=r)
{
if(Q[j].op==2) num[Q[j].c]+=get(Q[j].y);
if(Q[j].op==3) num[Q[j].c]-=get(Q[j].y);
tmp[++len]=Q[j++];
}
for(i=1;i<=st;i++) change(re[i],-1);
for(i=1;i<=len;i++) Q[l+i-1]=tmp[i];
}
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(q[i].op!=1) num[q[i].c]=0;Q[i]=q[i];}
cdq(l,r,L-1);
for(int i=l;i<=r;i++)
if(q[i].op!=1)
if(K[q[i].c]==num[q[i].c]) ans[q[i].c]=L;
else ans[q[i].c]=-1;
return;
}
int mid=(L+R)/2,l1=0,l2=0;
for(int i=l;i<=r;i++) {if(q[i].op!=1) num[q[i].c]=0;Q[i]=q[i];}
cdq(l,r,mid);
for(int i=l;i<=r;i++)
{
if(q[i].op==1)
{
if(q[i].c<=mid) q1[++l1]=q[i];
else q2[++l2]=q[i];
}
else
{
if(num[q[i].c]<K[q[i].c]) q1[++l1]=q[i];
else q2[++l2]=q[i];
}
}
tim++;
for(int i=l;i<=r;i++) if(q[i].op!=1&&num[q[i].c]<K[q[i].c]&&C[q[i].c]!=tim) K[q[i].c]-=num[q[i].c],C[q[i].c]=tim;
for(int i=1;i<=l1;i++) q[l+i-1]=q1[i];
for(int i=1;i<=l2;i++) q[l+i+l1-1]=q2[i];
solve(l,l+l1-1,L,mid);solve(l+l1,r,mid+1,R);
}
int main()
{
n=read();m=read();
for(int i=1;i<=m;i++)
{
int op;op=read();
if(op==1)
{
int x,y,c;x=read();y=read();c=read();
q[++cnt].op=1;q[cnt].x=x;q[cnt].y=y;q[cnt].c=c;
}
else
{
int x,y,x1,y1;qcnt++;x=read();y=read();x1=read();y1=read();K[qcnt]=read();
q[++cnt].op=2;q[cnt].x=x1;q[cnt].y=y1;q[cnt].c=qcnt;
q[++cnt].op=3;q[cnt].x=x-1;q[cnt].y=y1;q[cnt].c=qcnt;
q[++cnt].op=3;q[cnt].x=x1;q[cnt].y=y-1;q[cnt].c=qcnt;
q[++cnt].op=2;q[cnt].x=x-1;q[cnt].y=y-1;q[cnt].c=qcnt;
}
}
solve(1,cnt,1,inf);
for(int i=1;i<=qcnt;i++)
{
if(ans[i]==-1) printf("NAIVE!ORZzyz.\n");
else printf("%d\n",ans[i]);
}
}