二逼平衡树

码量稍微有大,不过思路清晰还是好写的

题目链接

外层线段树,内层平衡树

操作1 4 5就是在线段树上取出区间,然后平衡树内求答案,合并答案

修改也没什么好讲的,和上一题比较相似

操作2要二分答案,然后转化为1

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/*
Author: zxy_hhhh
date: 2018/12/03
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<cmath>
#define rep(x,a,b) for (int x=int(a);x<=(int)(b);x++)
#define drp(x,a,b) for (int x=int(a);x>=(int)(b);x--)
#define cross(x,a) for (int x=hd[a];x;x=nx[x])
#define ll long long
using namespace std;
inline ll rd()
{
ll _x=0;int _ch=getchar(),_f=1;
for(;!isdigit(_ch)&&(_ch!='-')&&(_ch!=EOF);_ch=getchar());
if (_ch=='-'){_f=0;_ch=getchar();}
for(;isdigit(_ch);_ch=getchar()) _x=_x*10+_ch-'0';
return _f?_x:-_x;
}
void write(ll _x){if (_x>=10) write(_x/10),putchar(_x%10+'0'); else putchar(_x+'0'); }
inline void wrt(ll _x,char _p){if (_x<0) putchar('-'),_x=-_x; write(_x); if (_p) putchar(_p);}
#define maxn 50005
#define inf 2147483647
#define mid ( (l + r) >> 1
int a[maxn], n, m;
namespace xtree {
struct node *nil;
struct node {
int sz, val, fix;
node *ls, *rs;
node(int x) : sz(1), val(x), fix(rand()) { ls = rs = nil; }
inline void update() { sz = ls->sz + rs->sz + 1; }
};
inline void init() {
nil = new node(0);
nil->ls = nil->rs = nil;
nil->sz = 0;
}
void split(node *now, int k, node *&x, node *&y, int op = 1) {
if (now == nil) {
x = y = nil;
return;
}
if (op == 1 ? now->val < k : now->ls->sz < k) {
x = now;
split(now->rs, (op == 1 ? k : k - now->ls->sz - 1), x->rs, y, op);
x->update();
} else {
y = now;
split(now->ls, k, x, y->ls, op);
y->update();
}
}
node *merge(node *x, node *y) {
if (x == nil) return y;
if (y == nil) return x;
if (x->fix < y->fix) {
x->rs = merge(x->rs, y);
return x->update(), x;
} else {
y->ls = merge(x, y->ls);
return y->update(), y;
}
}
inline void insert(node *&rt, int val) {
node *x, *y;
split(rt, val, x, y);
rt = merge(x, merge(new node(val), y));
}
inline void del(node *&rt, int val) {
node *x, *y, *z;
split(rt, val, x, y);
split(y, 1, y, z, 2);
rt = merge(x, z);
}
inline int pre(node *&rt, int val) {
node *x, *y, *z;
int ans;
split(rt, val, x, y), split(x, x->sz - 1, x, z, 2);
if (z == nil)
ans = -inf;
else
ans = (z->val);
rt = merge(x, merge(z, y));
return ans;
}
inline int nxt(node *&rt, int val) {
node *x, *y, *z;
int ans;
split(rt, val + 1, x, y), split(y, 1, y, z, 2);
if (y == nil)
ans = inf;
else
ans = y->val;
rt = merge(x, merge(y, z));
return ans;
}
inline int rank(node *&rt, int val) {
node *x, *y;
int ans;
split(rt, val, x, y);
if (x == nil)
ans = 0;
else
ans = x->sz;
rt = merge(x, y);
return ans;
}
} // namespace xtree
namespace ytree {
xtree::node *tr[maxn << 2];
inline int pre(int pos, int l, int r, int ql, int qr, int x) {
if (r < ql || l > qr) return -inf;
if (ql <= l && r <= qr) return x = xtree::pre(tr[pos], x);
return max(pre(pos << 1, l, mid, ql, qr, x),
pre(pos << 1 | 1, mid + 1, r, ql, qr, x));
}
inline int nxt(int pos, int l, int r, int ql, int qr, int x) {
if (r < ql || l > qr) return inf;
if (ql <= l && r <= qr) return x = xtree::nxt(tr[pos], x);
return min(nxt(pos << 1, l, mid, ql, qr, x),
nxt(pos << 1 | 1, mid + 1, r, ql, qr, x));
}
inline int rank(int pos, int l, int r, int ql, int qr, int x) {
if (r < ql || l > qr) return 0;
if (ql <= l && r <= qr) return x = xtree::rank(tr[pos], x);
return rank(pos << 1, l, mid, ql, qr, x) +
rank(pos << 1 | 1, mid + 1, r, ql, qr, x);
}
inline void change(int pos, int l, int r, int x, int v) {
xtree::del(tr[pos], a[x]), xtree::insert(tr[pos], v);
if (l == r) return;
if (x <= mid)
change(pos << 1, l, mid, x, v);
else
change(pos << 1 | 1, mid + 1, r, x, v);
}
inline int atrank(int L, int R, int k) {
int l = 0, r = 100000000, ans;
while (l <= r) {
if (rank(1, 1, n, L, R, mid) < k)
ans = mid, l = mid + 1;
else
r = mid - 1;
}
return ans;
}
void build(int pos, int l, int r) {
tr[pos] = xtree::nil;
rep(i, l, r) xtree::insert(tr[pos], a[i]);
if (l == r) return;
build(pos << 1, l, mid), build(pos << 1 | 1, mid + 1, r);
}
} // namespace ytree
int main() {
n = rd();
m = rd();
xtree::init();
rep(i, 1, n) a[i] = rd();
ytree::build(1, 1, n);
rep(i, 1, m) {
int op = rd();
if (op == 1) {
int l = rd(), r = rd(), x = rd();
wrt(ytree::rank(1, 1, n, l, r, x) + 1, '\n');
} else if (op == 2) {
int l = rd(), r = rd(), x = rd();
wrt(ytree::atrank(l, r, x), '\n');
} else if (op == 3) {
int x = rd(), k = rd();
ytree::change(1, 1, n, x, k);
a[x] = k;
} else if (op == 4) {
int l = rd(), r = rd(), x = rd();
wrt(ytree::pre(1, 1, n, l, r, x), '\n');
} else if (op == 5) {
int l = rd(), r = rd(), x = rd();
wrt(ytree::nxt(1, 1, n, l, r, x), '\n');
}
}
}
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