[ZJJOI2013]K大数查询 整体二分

[ZJJOI2013]K大数查询

链接

luogu

思路

整体二分。

代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll _=5e5+7;
ll read() {
	ll x=0,f=1;char s=getchar();
	for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
	for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
	return x*f;
}
ll n,m,ans[_];
struct OPT {ll opt,a,b,c,id;}Q[_],tmp1[_],tmp2[_];
namespace seg {
	#define ls rt<<1
	#define rs rt<<1|1
	struct node {
		ll l,r,lazy,siz;
		unsigned ll tot;
	}e[_<<2];
	void build(ll l,ll r,ll rt) {
		e[rt].l=l,e[rt].r=r,e[rt].siz=r-l+1;
		ll mid=(l+r)>>1;
		if(l==r) return;
		build(l,mid,ls);
		build(mid+1,r,rs);
	}
	void pushdown(ll rt) {
		if(e[rt].lazy) {
			e[ls].tot+=e[ls].siz*e[rt].lazy;
			e[rs].tot+=e[rs].siz*e[rt].lazy;
			e[ls].lazy+=e[rt].lazy;
			e[rs].lazy+=e[rt].lazy;
			e[rt].lazy=0;
		}
	}
	void modify(ll L,ll R,ll ad,ll rt) {
		if(L<=e[rt].l&&e[rt].r<=R) {
			e[rt].tot+=e[rt].siz*ad;
			e[rt].lazy+=ad;
			return;
		}
		ll mid=(e[rt].l+e[rt].r)>>1;
		pushdown(rt);
		if(L<=mid) modify(L,R,ad,ls);
		if(R>mid) modify(L,R,ad,rs);
		e[rt].tot=e[ls].tot+e[rs].tot;
	}
	unsigned ll query(ll L,ll R,ll rt) {
		if(L<=e[rt].l&&e[rt].r<=R) return e[rt].tot;
		ll mid=(e[rt].l+e[rt].r)>>1;
		unsigned ll ans=0;
		pushdown(rt);
		if(L<=mid) ans+=query(L,R,ls);
		if(R>mid) ans+=query(L,R,rs);
		return ans;
	}
}
void solve(ll l,ll r,ll vl,ll vr) {
	if(vl==vr||l>r) {
		for(ll i=l;i<=r;++i) ans[Q[i].id]=vl;
		return;
	}
	ll mid=(vl+vr)>>1,p=-1,q=-1;
	for(ll i=l;i<=r;++i) {
		if(Q[i].opt==1) {
			if(Q[i].c>mid) {
				seg::modify(Q[i].a,Q[i].b,1,1);
				tmp2[++q]=Q[i];
			} else tmp1[++p]=Q[i];
		} else {
			ll tmp=seg::query(Q[i].a,Q[i].b,1);
			if(Q[i].c<=tmp) tmp2[++q]=Q[i];
			else Q[i].c-=tmp,tmp1[++p]=Q[i];
		}
	}
	for(ll i=l;i<=r;++i)
		if(Q[i].opt==1&&Q[i].c>mid) seg::modify(Q[i].a,Q[i].b,-1,1);
	for(ll i=0;i<=p;++i) Q[l+i]=tmp1[i];
	for(ll i=0;i<=q;++i) Q[l+p+1+i]=tmp2[i];
	solve(l,l+p,vl,mid);
	solve(l+p+1,r,mid+1,vr);
}
int main() {
	n=read(),m=read();
	seg::build(1,n,1);
	ll cnt=0;
	for(ll i=1;i<=m;++i) {
		Q[i].opt=read();
		Q[i].a=read(),Q[i].b=read(),Q[i].c=read();
		if(Q[i].opt==2) Q[i].id=++cnt;
	}
	solve(1,m,-n,n);
	for(ll i=1;i<=cnt;++i)
		printf("%lld\n",ans[i]);
	return 0;
}