[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|1struct 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;
}