K-th Closest Distance

题目传送门

解题思路

二分答案+主席树

先建主席树，然后二分答案mid，在l和r的区间内查询[p-mid, p+mid]的范围内的数的个数，如果大于k则说明这个范围内存在第k小的数，r=mid，否则不存在，l=mid+1。

代码如下

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

inline int read(){
    int res = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        res = (res << 3) + (res << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -res : res;
}

const int N = 1000005;

int rt[N], cnt;
struct T{
    int l, r;
    int lch, rch;
    int sum;
}tree[N * 30];

inline int build(int l, int r)
{
    int p = ++cnt;
    tree[p].l = l;
    tree[p].r = r;
    tree[p].lch = tree[p].rch = tree[p].sum = 0;
    if(l == r)
        return p;
    int mid = (l + r) / 2;
    tree[p].lch = build(l, mid);
    tree[p].rch = build(mid + 1, r);
    return p;
}

inline int insert(int k, int x)
{
    int p = ++cnt;
    tree[p].l = tree[k].l;
    tree[p].r = tree[k].r;
    tree[p].lch = tree[k].lch;
    tree[p].rch = tree[k].rch;
    tree[p].sum = tree[k].sum + 1;
    if(tree[k].l == tree[k].r)
        return p;
    int mid = (tree[k].l + tree[k].r) / 2;
    if(x <= mid)
        tree[p].lch = insert(tree[k].lch, x);
    else
        tree[p].rch = insert(tree[k].rch, x);
    return p;
}

int query(int k1, int k2, int l, int r)
{
    if(tree[k1].l >= l && tree[k1].r <= r)
        return tree[k2].sum - tree[k1].sum;
    int mid = (tree[k1].l + tree[k1].r) / 2;
    int m1 = 0, m2 = 0;
    if(l <= mid)
        m1 = query(tree[k1].lch, tree[k2].lch, l, r);
    if(r > mid)
        m2 = query(tree[k1].rch, tree[k2].rch, l, r);
    return m1 + m2;
}

int main()
{
    int _;
    scanf("%d", &_);
    while(_ --){
        int n, m;
        scanf("%d%d", &n, &m);
        rt[0] = build(1, 1e6);
        for(int i = 1; i <= n; i ++){
            int x = read();
            rt[i] = insert(rt[i -1], x);
        }
        int x = 0;
        for(int i = 1; i <= m; i ++){
            int l, r, p, k;
            l = read(), r = read(), p = read(), k = read();
            l ^= x, r ^= x, p ^= x, k ^= x;
            int ll = 0, rr = 1e6;
            int mid = (ll + rr) / 2;
            while(ll < rr){
                if(query(rt[l - 1], rt[r], max(1, p - mid), min(p + mid, 1000000)) >= k)
                    rr = mid;
                else
                    ll = mid + 1;
                mid = (ll + rr) / 2;
            }
            x = mid;
            printf("%d\n", x);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/whisperlzw/p/11280421.html