hdu6621-- Chairman tree

K-th Closest Distance

The meaning of problems: a given length of \ (n-\) is the number of columns, m prior queries, each query to \ (L, R & lt, P, K \) , for Q \ (l <= i <= r \) , \ (| PA [I] | \) is a small value of k. Forced data online.

Solution: binary answer ANS, with Chairman query tree \ (a [l] \) to \ (a [r] \) between \ ([p-ans, p + ans] \) number, the number is k is the answer.

Due to mandatory online, so the tree can not be built Chairman discrete. Direct contribution of dynamic space.

#include<bits/stdc++.h>
using namespace std;
const int maxn=100005;
int n,m;
const int N=1e6;
struct node{
    int ls,rs,sum;
};
node tree[maxn*22];
int rt[maxn],tot=0;
int update(int l,int r,int x,int pre){
    int now=++tot;
    tree[now]=tree[pre];
    if(l==r){
        tree[now].sum++; return now;
    }
    int mid=(l+r)>>1;
    if(x>mid) tree[now].rs=update(mid+1,r,x,tree[pre].rs);
    else tree[now].ls=update(l,mid,x,tree[pre].ls);
    tree[now].sum++;
    return now;
}
int ask(int l,int r,int il,int ir,int root1,int root2){
    if(r<il||l>ir) return 0;
    if(l>=il&&r<=ir) return tree[root2].sum-tree[root1].sum;
    int mid=(l+r)>>1;
    return ask(l,mid,il,ir,tree[root1].ls,tree[root2].ls)+ask(mid+1,r,il,ir,tree[root1].rs,tree[root2].rs);
}
int jud(int len,int p,int rt1,int rt2){
    int l=max(1,p-len),r=min(N,p+len);
    return ask(1,N,l,r,rt1,rt2);
}
int solve(int l,int r,int p,int k){
    int rt1=rt[l-1],rt2=rt[r];
    int ll=0,rr=N;
    while(ll<rr){
        int mid=(ll+rr)>>1;
        if(jud(mid,p,rt1,rt2)>=k) rr=mid;
        else ll=mid+1;
    }
    return ll;
}
int main(){
    int T;
    cin >> T;
    while(T--){
        tot=0;
        tree[0].sum=tree[0].ls=tree[0].rs=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) {
            int x;
            scanf("%d",&x);
            rt[i]=update(1,N,x,rt[i-1]);
        }
        int las=0;
        while(m--){
            int l,r,p,k;
            scanf("%d%d%d%d",&l,&r,&p,&k);
            l^=las; r^=las; p^=las; k^=las;
            printf("%d\n",las=solve(l,r,p,k));
        }
    }
    return 0;
}