洛谷 [P3834] 可持久化线段树（主席树）

主席树可以存储线段树的历史状态，空间消耗很大，一般开45n即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <queue>
#define lson l, mid
#define rson mid+1, r
#define ll long long
using namespace std;
const int MAXN = 200005;
ll init() {
    ll rv = 0, fh = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-' ) fh = -1;
        c = getchar();
    }
    while(c >= '0' && c <='9') {
        rv = (rv<<1) + (rv<<3) + c - '0';
        c = getchar();
    }
    return fh * rv;
}
ll n, m, num[MAXN],tmp[MAXN];
struct LTSGT {
    ll rot[MAXN], sum[MAXN*45], lpos[MAXN*45], rpos[MAXN*45];
    int cnt = 0;
    int build(int l, int r) {
        int rt = ++cnt;
        if(l == r) {
            sum[rt] = 0;
            return rt;
        }
        int mid = (l + r) >>1;
        lpos[rt] = build(l, mid);
        rpos[rt] = build(mid + 1, r);
        return rt;
    }
    int Update(int pre, int l, int r, int x) {
        int rt = ++cnt;
        if(l == r) {
            sum[rt] = sum[pre] + 1;
            return rt;
        }
        int mid = (l + r) >>1;
        lpos[rt] = lpos[pre]; rpos[rt] = rpos[pre]; sum[rt] = sum[pre] + 1;
        if(x <= mid) lpos[rt] = Update(lpos[pre], l, mid, x);
        else rpos[rt] = Update(rpos[pre], mid + 1, r, x);
        return rt;
    }
    int Query(int aa, int bb, int l, int r, int k) {
        if(l == r) return l;
        int mid = (l + r) >>1;
        int x = sum[lpos[bb]] - sum[lpos[aa]];
        if(x >= k) return Query(lpos[aa], lpos[bb], l, mid, k);
        else return Query(rpos[aa], rpos[bb], mid + 1, r, k - x); //注意。这里要k-x
    }
    void print(int rt){
        cout<<sum[rt]<<endl;
        if(lpos[rt]) print(lpos[rt]);
        if(rpos[rt]) print(rpos[rt]);
    }
}ltsgt;
int main() {
    freopen("in.txt", "r", stdin);
    n = init(); m = init();
    for(int i = 1 ; i <= n ; i++) {
        num[i] = init();
        tmp[i] = num[i];
    }
    sort(tmp + 1, tmp + n + 1);
    ltsgt.rot[0] = ltsgt.build(1, n);
    for(int i = 1 ; i <= n ; i++) {
        int t = lower_bound(tmp + 1, tmp + n + 1, num[i]) - tmp;
        ltsgt.rot[i] = ltsgt.Update(ltsgt.rot[i-1], 1, n ,t);
    }
    //ltsgt.print(ltsgt.rot[4]);
    for(int i = 1 ; i <= m ; i++) {
        int aa = init(), bb = init(), k = init();
        printf("%lld\n", tmp[ltsgt.Query(ltsgt.rot[aa-1], ltsgt.rot[bb], 1, n, k)]); //注意这里是aa-1
    }
    fclose(stdin);
    return 0;
}

原文地址：https://www.cnblogs.com/Mr-WolframsMgcBox/p/8452287.html