可持久化线段树

题目

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 100009;
int n, q, cnt, sz, Max[MAXN * 20], T[MAXN * 20], ls[MAXN * 20], rs[MAXN * 20];
void pushup(int rt) { Max[rt] = max(Max[ls[rt]], Max[rs[rt]]); }
void build(int &rt, int l, int r)
{
    rt = ++sz;
    if (l == r) {
        scanf("%d", &Max[rt]);
        return;
    }
    int mid = (l + r) >> 1;
    build(ls[rt], l, mid);
    build(rs[rt], mid + 1, r);
    pushup(rt);    
}
int query(int rt, int l, int r, int ql, int qr)
{
    if (ql == l && qr == r) return Max[rt];
    int mid = (l + r) >> 1;
    if (qr <= mid) return  query(ls[rt], l, mid, ql, qr);
    else if (ql > mid) return query(rs[rt], mid + 1, r, ql, qr);
    else return max(query(ls[rt], l, mid, ql, mid), query(rs[rt], mid + 1, r, mid + 1, qr));
}
void update(int &rt, int fa, int l, int r, int id, int v)
{
    rt = ++sz;
    Max[rt] = Max[fa];
    if (l == r) {
        Max[rt] = v;
        return;
    }
    ls[rt] = ls[fa];
    rs[rt] = rs[fa];
    int mid = (l + r) >> 1;
    if (id <= mid) update(ls[rt], ls[fa], l, mid, id, v);
    else update(rs[rt], rs[fa], mid + 1, r, id, v);
    pushup(rt);
}
int main()
{
    sz = cnt = 0;
    scanf("%d %d", &n, &q);
    build(T[++cnt], 1, n);
    while (q--) {
        int op, x, y, z;
        scanf("%d %d %d %d", &op, &x, &y, &z);
        if (op == 0) {
            int ans = query(T[x], 1, n, y, z);
            printf("%d
", ans);
        }
        else {
            update(T[++cnt], T[x], 1, n, y, z);
        }
    }
    return 0;
}

题目

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 100009;
int t, n, m, a[MAXN], b[MAXN], sz, sum[MAXN * 20], ls[MAXN * 20], rs[MAXN * 20], T[MAXN * 20];
void pushup(int rt) { sum[rt] = sum[ls[rt]] + sum[rs[rt]]; }
/*
void build(int &rt, int l, int r)
{
    rt = ++sz;
    sum[rt] = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(ls[rt], l, mid);
    build(rs[rt], mid + 1, r);
}
*/
void update(int &rt, int fa, int l, int r, int id)
{
    rt = ++sz;
    sum[rt] = sum[fa] + 1;
    if (l == r) return;
    ls[rt] = ls[fa];
    rs[rt] = rs[fa];
    int mid = (l + r) >> 1;
    if (id <= mid) update(ls[rt], ls[fa], l, mid, id);
    else update(rs[rt], rs[fa], mid + 1, r, id);
    pushup(rt);
}
int query(int u, int v, int l, int r, int z)
{
    if (l == r) return l;
    int tmp = sum[ls[v]] - sum[ls[u]];
    int mid = (l + r) >> 1;
    if (z <= tmp) {
        return query(ls[u], ls[v], l, mid, z);
    }
    else return query(rs[u], rs[v], mid + 1, r, z - tmp);
}
int main()
{
    scanf("%d", &t);
    while (t--) {
        sz = 0;
        scanf("%d %d", &n, &m);
        memset(sum, 0, sizeof(sum));
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
            b[i] = a[i];
        }
        sort(a + 1, a + 1 + n);
        int nn = unique(a + 1, a + 1 + n) - a - 1;
        //build(T[0], 1, nn);
        for (int i = 1; i <= n; ++i) {
            int id = lower_bound(a + 1, a + 1 + nn, b[i]) - a;
            update(T[i], T[i - 1], 1, nn, id);
        }
        for (int i = 1; i <= m; ++i) {
            int x, y, z;
            scanf("%d %d %d", &x, &y, &z);
            int id = query(T[x - 1], T[y], 1, nn, z);
            printf("%d
", a[id]);
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/--ZHIYUAN/p/9043343.html