K-th Closest Distance

Posted 2020-11-22 2462478392lee

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6621

题意：

对于每个查询，都会给您一个间隔[L，R]和两个数字p和K。您的目标是找到p和aL，aL+1，…，aR之间的第K个最近距离。

p和ai之间的距离等于| p-ai |。

思路：直接建树，二分绝对值ans，让主席树查找[p-ans,p+ans]之间的数是不是又k个，如果是，那么答案就是ans。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
#include<vector>
#include<queue>
#include<cmath>
#define ll long long
using namespace std;
const int N=1e5+10;
const int M=1e6;
struct node
{
    int l, r;
    int val;
}tree[N*55];
int root[N], tot;
int n,q,a[N];
int update(int pre, int pl, int pr, int val)
{
    int cur = ++tot;
    tree[cur] = tree[pre];
    tree[cur].val ++;
    if(pl == pr)
        return cur;
    int mid = (pl + pr) >> 1;
    if(val <= mid)
        tree[cur].l = update(tree[pre].l, pl, mid, val);
    else
        tree[cur].r = update(tree[pre].r, mid + 1, pr, val);
    return cur;
}
int k, cnt;
int query(int pl, int pr, int l, int r, int rt, int lt)
{
    if(pl <= l && r <= pr)
    {
        return tree[rt].val - tree[lt].val;
    }
    int mid = (l + r) >> 1;
    int res = 0;
    if(pl <= mid)
        res += query(pl, pr, l, mid, tree[rt].l, tree[lt].l);
    if(pr > mid)
        res += query(pl, pr, mid + 1, r, tree[rt].r, tree[lt].r);
    return res;
}
int main()
{
    int T, ans;
    int l, r, p;
    int pl, pr, mid;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &q);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            root[i] = update(root[i - 1], 1, M, a[i]);
        }
        ans = 0;
        while(q--)
        {
            scanf("%d%d%d%d", &l, &r, &p, &k);
            l = l ^ ans;
            r = r ^ ans;
            p = p ^ ans;
            k = k ^ ans;
            pl = 0, pr = M;
            while(pl<=pr)
            {
                mid = (pl + pr) >> 1;
                if(query(max(1, p - mid), min(M, p + mid), 1, M, root[r], root[l - 1]) >= k)
                {
                    ans = mid;
                    pr = mid - 1;
                }
                else
                    pl = mid + 1;
            }
            printf("%d
", ans);
        }
    }
    return 0;
}