洛谷 - P3332 [ZJOI2013] K大数查询

传送门：洛谷 - P3332

思路分析

一般查询第K大数这类问题，可以在值域线段树上快速解决，所以这道题外层为值域线段树，内层套下标线段树
因为涉及到区间修改，动态开点的话pushdown的时候要新建，所以采用标记永久化，这样就不用pushdown了

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define int long long
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 5e4+10;

int n,q;

//下标
struct node1{
    ll sum;
    int tag,ls,rs;
}xtree[N * 400];

int cntx;
#define ls xtree[x].ls
#define rs xtree[x].rs
void change(int &x,int l,int r,int L,int R){
    if(!x) x=++cntx;
    xtree[x].sum+=R-L+1;
    if(l==L && r==R){
        xtree[x].tag++;
        return ;
    }
    int mid=l+r>>1;
    if(R<=mid) change(ls,l,mid,L,R);
    else if(L>mid) change(rs,mid+1,r,L,R);
    else{
        change(ls,l,mid,L,mid);
        change(rs,mid+1,r,mid+1,R);
    }
    return ;
}

ll query(int &x,int l,int r,int L,int R,int tag){
    if(!x) return 1LL*tag*(R-L+1);
    if(l==L && r==R){
        return xtree[x].sum+1LL*tag*(R-L+1);
    }
    int mid=l+r>>1;
    tag+=xtree[x].tag;
    if(R<=mid) return query(ls,l,mid,L,R,tag);
    else if(L>mid) return query(rs,mid+1,r,L,R,tag);
    else return query(ls,l,mid,L,mid,tag) + query(rs,mid+1,r,mid+1,R,tag);
}
#undef ls
#undef rs

//值域
struct node2{
    int root,l,r,ls,rs;
}vtree[N * 4];

int root,cntv;

#define ls vtree[x].ls
#define rs vtree[x].rs
void build(int &x,int l,int r){
    if(!x) x=++cntv;
    vtree[x].l=l;
    vtree[x].r=r;
    if(l==r) return ;
    int mid=l+r>>1;
    build(ls,l,mid);
    build(rs,mid+1,r);
} 

void modify(int x,int L,int R,int pos){
    change(vtree[x].root,1,n,L,R);
    if(vtree[x].l == vtree[x].r){
        return ;
    }
    int mid=vtree[x].l+vtree[x].r>>1;
    if(pos<=mid) modify(ls,L,R,pos);
    else modify(rs,L,R,pos);
}

int ask(int x,int L,int R,int k){
    if(vtree[x].l==vtree[x].r) return vtree[x].l;
    int tmp=query(vtree[rs].root,1,n,L,R,0);
    if(k<=tmp) return ask(rs,L,R,k);
    else return ask(ls,L,R,k-tmp);
}
#undef ls
#undef rs

signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    read(n,q);
    build(root,-n,n);
    while(q--){
        int op,l,r;
        ll x;
        read(op,l,r,x);
        if(op==1) modify(root,l,r,x);
        else printf("%lld\n",ask(root,l,r,x));
    }



    return 0;
}