2019杭电多校第三场 1008 K-th Closest Distance

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

考虑主席树，我们先将所有值离散化之后建主席树。对于每个查询\(s,t,p,k\)
我们考虑二分一个值\(mid\)，考虑当前区间内，\([p-mid, p+mid]\)的值有多少个，很显然这是符合单调性的，那么我们只需要每次判断即可。时间复杂度\(O(nlog^2n)\)

#include <bits/stdc++.h>
#define pii pair<int, int>
#define pil pair<int, long long>
#define pll pair<long long, long long>
#define lowbit(x) ((x)&(-x))
#define mem(i, a) memset(i, a, sizeof(i))
#define sqr(x) ((x)*(x))
#define all(x) x.begin(),x.end()
#define ls (k << 1)
#define rs (k << 1 | 1)
using namespace std;
typedef long long ll;
template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for(; ch > '9' || ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

const int INF = 0x3f3f3f3f;
const int N = 1e5 + 5;
int T[N << 5],L[N << 5],R[N << 5],sum[N << 5];
int n,m,cnt;
int a[N],id[N];
int update(int pre, int l, int r, int x) {
    int rt = ++cnt;
    L[rt] = L[pre]; R[rt] = R[pre]; sum[rt] = sum[pre] + 1;
    //cout << rt << " " << sum[rt] << "\n";
    int mid = l + r >> 1;
    if(l < r) {
        if(x <= mid) L[rt] = update(L[pre], l, mid, x);
        else R[rt] = update(R[pre], mid + 1, r, x);
    }
    return rt;
}
int query(int u, int v, int l, int r, int s, int t)  {
    int x = sum[v] - sum[u];
   // cout << l << " " << r << " " << s << " " << t << " " << L[v] << " " << R[v] <<  "\n";
    int ans = 0;
    if(s <= l && t >= r) {
        return x;
    }
    int mid = l + r >> 1;
    if(s <= mid) ans += query(L[u], L[v], l, mid, s, t);
    if(t > mid) ans += query(R[u], R[v], mid + 1, r, s, t);
    return ans;
}
int main() {
#ifdef INCTRY
    freopen("input.txt", "rt", stdin);
#endif
    int t;
    cin >> t;
    while(t--) {
        read(n); read(m); cnt = 0;
        vector<int> v;

        for(int i = 1; i <= n; i++) {
            read(a[i]);
            v.push_back(a[i]);
        }
        sort(all(v)); v.erase(unique(all(v)), v.end());
        for(int i = 1; i <= n; i++) {
            id[i] = lower_bound(all(v), a[i]) - v.begin() + 1;
        }
        int nall = v.size();
        T[0] = 0;

        for(int i = 1; i <= n; i++) {
            T[i] = update(T[i - 1], 1, nall, id[i]);
        }
        //cout << n << " " << m <<"\n";
        int prv = 0;
        int ans;
        //cout << T[5] << " " << L[T[5]] << " " << R[T[5]] << "\n";
       //cout << query(T[0], T[5], 1, nall, 1, 4);
         for(int i = 1; i <= m; i++) {
            int s,t,p,k;
            //cout << i << " " << m << "\n";
            read(s); read(t); read(p); read(k);
            s ^= prv; t ^= prv; p ^= prv; k ^= prv;
            int l = 0, r = 1e8;
            while(l <= r) {
                int mid = l + r >> 1;
                int p1 = lower_bound(all(v), p - mid) - v.begin() + 1;
                int p2 = upper_bound(all(v), p + mid) - v.begin();
                //cout << p1 << " "  << p2 << " " << p - mid << " " << p + mid << " ";
                int tot = query(T[s-1], T[t], 1, nall, p1, p2); //cout << tot <<"\n";
                if(tot >= k) ans = mid, r = mid - 1;
                else l = mid + 1;
            }
            cout << ans << "\n";
            prv = ans;
        }
    }

#ifdef INCTRY
    cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
    return 0;
}

原文地址：https://www.cnblogs.com/inctry/p/11287658.html