主席树

一种可持续化的数据结构

可用于计算区间第(k)大

主席树详解

[模板]可持久化线段树1(主席树)

题目链接

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 200100
int n, q;
int a[maxn], b[maxn], root[maxn];
int cnt = 0;
int lz[maxn << 5], rz[maxn << 5], sum[maxn << 5];
int build(int l, int r)
{
    int rt = ++cnt;
    sum[rt] = 0;
    int mid = (l + r) >> 1;
    if(l == r) return rt;
    lz[rt] = build(l, mid);
    rz[rt] = build(mid + 1, r);
    return rt;
}
int update(int pre, int l, int r, int x)
{
    int rt = ++cnt;
    lz[rt] = lz[pre]; rz[rt] = rz[pre]; sum[rt] = sum[pre] + 1;
    if(l == r) return rt;
    int mid = (l + r) >> 1;
    if(mid >= x) lz[rt] = update(lz[pre], l, mid, x);
    if(mid < x) rz[rt] = update(rz[pre], mid + 1, r, x);
    return rt;
}
int query(int u, int v, int l, int r, int k)
{
    int x = sum[lz[v]] - sum[lz[u]];
    int mid = (l + r) >> 1;
    if(l == r) return l;
    if(k <= x) return query(lz[u], lz[v], l, mid, k);
    else return query(rz[u], rz[v], mid + 1, r, k - x);
}
int main()
{
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
    sort(b + 1, b + n + 1);
    int m = unique(b + 1, b + n + 1) - (b + 1);
    root[0] = build(1, m);
    for(int i = 1; i <= n; i++)
    {
        int x = lower_bound(b + 1, b + m + 1, a[i]) - b;
        root[i] = update(root[i - 1], 1, m, x);
    }
    for(int i = 1; i <= q; i++)
    {
        int l, r, k;
        scanf("%d%d%d", &l, &r, &k);
        int t = query(root[l - 1], root[r], 1, m, k);
        printf("%d
", b[t]);
    }
    return 0;
}

[模板]可持久化数组

题目链接

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 1000100
int n, q;
int cnt = 0;
int a[maxn], b[maxn];
int sum[maxn << 5], root[maxn << 5];
int lz[maxn << 5], rz[maxn << 5];
int build(int l, int r)
{
    int rt = ++cnt;
    if(l == r)
    {
        sum[rt] = a[l];
        return rt;
    }
    int mid = (l + r) >> 1;
    lz[rt] = build(l, mid);
    rz[rt] = build(mid + 1, r);
    return rt;
}
int update(int pre, int l, int r, int x, int v)
{
    int rt = ++cnt;
    lz[rt] = lz[pre]; rz[rt] = rz[pre];
    if(l == r)
    {
        sum[rt] = v; return rt;
    }
    int mid = (l + r) >> 1;
    if(mid >= x) lz[rt] = update(lz[pre], l, mid, x, v);
    if(mid < x) rz[rt] = update(rz[pre], mid + 1, r, x, v);
    return rt;
}
int ans = 0;
int query(int k, int l, int r, int x)
{
    if(l == r)
    {
        return sum[k];
    }
    int mid = (l + r) >> 1;
    if(mid >= x) return query(lz[k], l, mid, x);
    if(mid < x) return query(rz[k], mid + 1, r, x);
}
int main()
{
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    root[0] = build(1, n);
    for(int i = 1; i <= q; i++)
    {
        int pre; scanf("%d", &pre);
        int op; scanf("%d", &op);
        if(op == 1)
        {
            int x, v; scanf("%d%d", &x, &v);
            root[i] = update(root[pre], 1, n, x, v);
        }
        if(op == 2)
        {
            int x; scanf("%d", &x);
            root[i] = root[pre];
            printf("%d
", query(root[pre], 1, n, x));
        }
    }
    return 0;
}