洛谷 P3834 【模板】可持久化线段树 2（主席树）

算法数据结构还是得时常复习，达到半夜被叫醒也能流畅写完的境界才算是学通透了。

传送门

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using p = pair<int, int>;
const double pi(acos(-1));
const int inf(0x3f3f3f3f);
const int mod(1e9 + 7);
const int maxn(2e5 + 10);
int cnt, a[maxn], root[maxn];
vector<int> nums;

struct node {
    int l, r, sum;
} tree[maxn * 40];

template<typename T = int>
inline const T read()
{
    T x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 3) + (x << 1) + ch - '0';
        ch = getchar();
    }
    return x * f;
}

template<typename T>
inline void write(T x, bool ln)
{
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x / 10, false);
    putchar(x % 10 + '0');
    if (ln) putchar(10);
}

inline int getId(int val)
{
    return int(lower_bound(nums.begin(), nums.end(), val) - nums.begin()) + 1;
}

void insert(int l, int r, int pos, int pre, int& cur)
{
    cur = ++cnt;
    tree[cur] = tree[pre];
    ++tree[cur].sum;
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (pos <= mid) {
        insert(l, mid, pos, tree[pre].l, tree[cur].l);
    } else {
        insert(mid + 1, r, pos, tree[pre].r, tree[cur].r);
    }
}

int query(int l, int r, int k, int pre, int cur)
{
    if (l == r) return l;
    int mid = (l + r) >> 1;
    int dif = tree[tree[cur].l].sum - tree[tree[pre].l].sum;
    if (k <= dif) {
        return query(l, mid, k, tree[pre].l, tree[cur].l);
    }
    return query(mid + 1, r, k - dif, tree[pre].r, tree[cur].r);
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("input.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    int n = read(), m = read();
    for (int i = 1; i <= n; ++i) {
        nums.push_back(a[i] = read());
    }
    sort(nums.begin(), nums.end());
    nums.erase(unique(nums.begin(), nums.end()), nums.end());
    for (int i = 1; i <= n; ++i) {
        int pos = getId(a[i]);
        insert(1, (int)nums.size(), pos, root[i - 1], root[i]);
    }
    while (m--) {
        int l = read(), r = read(), k = read();
        write(nums[query(1, (int)nums.size(), k, root[l - 1], root[r]) - 1], true);
    }
    return 0;
}