[Luogu] 可持久化线段树 1（主席树）

Posted 2020-10-24 xayata

https://www.luogu.org/problemnew/show/P3834

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;
const int maxn = 2e5 + 10;

#define RR freopen("gg.in", "r", stdin)

int n, m;
int cnt;

struct node {
    int L, R;//分别指向左右子树
    int sum;//该节点对应区间数的个数
    node() {
        sum = 0;
    }
} Tree[maxn * 20];

struct value {
    int x;//值的大小
    int id;//离散之前在原数组中的位置
} Value[maxn];

bool cmp(value v1, value v2) {
    return v1.x < v2.x;
}

int root[maxn];//多颗线段树的根节点
int rank[maxn];//原数组离散之后的数组

void init() {
    cnt = 1;
    root[0] = 0;
    Tree[0].L = Tree[0].R = Tree[0].sum = 0;
}

void update(int num, int &rt, int l, int r) {
    Tree[cnt++] = Tree[rt];//bu存在Tree[0] 
    rt = cnt - 1;
    Tree[rt].sum++;
    if(l == r) return;
    int mid = (l + r)>>1;
    if(num <= mid) update(num, Tree[rt].L, l, mid);
    else update(num, Tree[rt].R, mid + 1, r);
}

int query(int i, int j, int k, int l, int r) {
    int d = Tree[Tree[j].L].sum - Tree[Tree[i].L].sum;
    if(l == r) return l;
    int mid = (l + r)>>1;
    if(k <= d) return query(Tree[i].L, Tree[j].L, k, l, mid);
    else return query(Tree[i].R, Tree[j].R, k - d, mid + 1, r);
}

int main() {
    RR;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &Value[i].x);
        Value[i].id = i;
    }
    //进行离散化
    sort(Value + 1, Value + n + 1, cmp);
    for(int i = 1; i <= n; i++) rank[Value[i].id] = i;
    init();
    for(int i = 1; i <= n; i++) {
        root[i] = root[i - 1];
    //    cout << root[i] << endl;
        update(rank[i], root[i], 1, n);//update(第i个数的排名，)
    }
    int left, right, k;
    for(int i = 1; i <= m; i++) {
        scanf("%d%d%d", &left, &right, &k);
        printf("%d\n", Value[query(root[left - 1], root[right], k, 1, n)].x);
    }
    return 0;
}