Cutting Bamboos

可持久线段树

https://ac.nowcoder.com/acm/problem/52172

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<math.h>
#include<queue>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
#define me0(a) memset(a,0,sizeof(a))
#define me1(a) memset(a,-1,sizeof(a))
using namespace std;
 
const int maxn = 4e6 + 100;
const int INF = 0x3f3f3f3f;
typedef long long LL;
 
void read(LL& val) { int x = 0; int bz = 1; char c; for (c = getchar(); (c<'0' || c>'9') && c != '-'; c = getchar()); if (c == '-') { bz = -1; c = getchar(); }for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48; val = x * bz; }
 
LL n, q, a[maxn], b[maxn], sz, tot, root[maxn];
LL ls[maxn], rs[maxn], rt[maxn], val[maxn];
LL s[maxn];//s表示总和
LL sum[maxn];
 
void build(LL& o, int L, int R) {
    o = ++tot;
    val[o] = 0; s[o] = 0;
    if (L == R) return;
    int m = (L + R) >> 1;
    build(ls[o], L, m);
    build(rs[o], m + 1, R);
}
 
void update(LL& o, int L, int R, int last, int p) {//last 指的是上一个root的结点编号
    o = ++tot;
    ls[o] = ls[last];//
    rs[o] = rs[last];
    s[o] = s[last] + p;
    val[o] = val[last] + 1;//表示当前结点的重量增加1
    if (L == R) return;
    int m = (L + R) >> 1;
    if (p <= m) update(ls[o], L, m, ls[last], p);//根据p值判断插入左子树还是右子树中
    else update(rs[o], m + 1, R, rs[last], p);
}
 
int query(int ss, int tt, int L, int R, int k) {//区间起点、终点，查询的第k小
    if (L == R) return L;
    int m = (L + R) >> 1;
    int cnt = val[ls[tt]] - val[ls[ss]];//结点的左子树相减，表示这个期间左子树上增加了多少个新的值，然后常规操作
    if (k <= cnt) return query(ls[ss], ls[tt], L, m, k);//说明第k小在当前m的左边
    else return query(rs[ss], rs[tt], m + 1, R, k - cnt);
 
}
 
LL query2(int ss, int tt, int L, int R, int k) {//查询前k小的和
    if (L == R)  {
        return (LL)k * L;
    }
    int m = (L + R) >> 1;
 
    int cnt = val[ls[tt]] - val[ls[ss]];
    if (k > cnt) {
        return s[ls[tt]] - s[ls[ss]] + query2(rs[ss], rs[tt], m + 1, R, k - cnt);
    }
    else {
        return query2(ls[ss], ls[tt], L, m, k);
    }
 
}
 
void work() {
    LL ql, qr, x, y;
    read(ql); read(qr); read(x); read(y);
    double dpart = double(sum[qr] - sum[ql - 1]) / y * x;//切掉部分的长度和
    LL L = 0, R = qr - ql + 1, m = 0, res = 0;
    LL num = qr - ql + 1;//总共num根竹子
    LL Kth = a[ql], Ksum = 0, K = 0;
    while (L <= R) {
        m = (L + R) >> 1;
        Kth = query(root[ql - 1], root[qr], 1, sz, m);//找到m小,表示切到了第m小
        Ksum = query2(root[ql - 1], root[qr], 1, sz, m);//前m小
        Ksum = sum[qr] - sum[ql - 1] - Ksum - (num - m) * Kth;//切掉的部分
 
        if ((double)Ksum - 0.00000001 >= dpart) {
            res = Kth;
            K = m;
            L = m + 1;
        }
        else {
            R = m - 1;
        }
    }
    double ans = Kth + ((double)Ksum - dpart) / (num - K);
    printf("%.9lf\n", ans);
     
}
 
int main() {
    //freopen("in.txt", "r", stdin);
    read(n); read(q);
    fo(i, 1, n) read(a[i]), sum[i] = sum[i - 1] + a[i], sz = max(sz, a[i]);//这里a[i]是连续的，不连续的需要另外做处理
    tot = 0;
    build(root[0], 1, sz);//首先建立一个空树
 
    fo(i, 1, n) update(root[i], 1, sz, root[i - 1], a[i]);//按输入顺序依次插入到相应的结点中去，并更改途中结点的信息（此时该结点有多少个子节点）
 
    while (q--) work();
    return 0;
}

 

https://cloud.tencent.com/developer/article/1421786