bzoj3110: [Zjoi2013]K大數查詢(權值線段樹套區間線段樹)
阿新 • • 發佈:2018-12-17
題目
內層線段樹維護權值k在[l,r]內出現次數
#include<bits/stdc++.h>
using namespace std;
typedef unsigned int ll;
#define mid ((l+r)>>1)
const int N=200002,M=20000002;
int n,m,op,x,y,z,lz[M],L[M],R[M],cnt,rt[N];
ll sum[M];
inline char gc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2= (p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
int x=0,fl=1;char ch=gc();
for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
return x*fl;
}
inline void wri(int a){if(a>=10)wri(a/10);putchar (a%10|48);}
inline void wln(int a){if(a<0)a=-a,putchar('-');wri(a),puts("");}
void down(int t,int l,int r){
if (lz[t]){
if (!L[t]) L[t]=++cnt;
if (!R[t]) R[t]=++cnt;
sum[L[t]]+=(mid-l+1)*lz[t];
sum[R[t]]+=(r-mid)*lz[t];
lz[L[t]]+=lz[t];
lz[R[t]]+= lz[t];
lz[t]=0;
}
}
void update(int &t,int l,int r,int x,int y){
if (!t) t=++cnt;
if (x<=l && r<=y){
sum[t]+=r-l+1;
lz[t]++;
return;
}
down(t,l,r);
if (x<=mid) update(L[t],l,mid,x,y);
if (mid<y) update(R[t],mid+1,r,x,y);
sum[t]=sum[L[t]]+sum[R[t]];
}
ll query(int t,int l,int r,int x,int y){
if (!t) return 0;
if (x<=l && r<=y) return sum[t];
down(t,l,r);
ll ans=0;
if (x<=mid) ans+=query(L[t],l,mid,x,y);
if (mid<y) ans+=query(R[t],mid+1,r,x,y);
return ans;
}
void ins(){
int t=1,l=1,r=n;
while (l<r){
update(rt[t],1,n,x,y);
if (z<=mid) r=mid,t<<=1;
else l=mid+1,t=t<<1|1;
}
update(rt[t],1,n,x,y);
}
int solve(){
int t=1,l=1,r=n;
while (l<r){
ll p=query(rt[t<<1],1,n,x,y);
if (z<=p) r=mid,t<<=1;
else l=mid+1,t=t<<1|1,z-=p;
}
return l;
}
int main(){
n=rd(),m=rd();
for (;m--;){
op=rd(),x=rd(),y=rd(),z=rd();
if (op==1) z=n+1-z,ins();
else wln(n+1-solve());
}
}