标签：
http://acm.hdu.edu.cn/showproblem.php?pid=5412
There are $N$ boys in CodeLand.
Boy i has his coding skill $A_{i}$.
CRB wants to know who has the suitable coding skill.
So you should treat the following two types of queries.
Query 1: 1 $l\ v$
The coding skill of Boy l has changed to $v$.
Query 2: 2 l $r\ k$
This is a report query which asks the $k-th$ smallest value of coding skill between Boy $l$ and Boy $r$(both inclusive).
There are multiple test cases.
The first line contains a single integer $N$.
Next line contains $N$ space separated integers $A_{1}, A_{2}, …, A_{N}$, where $A_{i}$ denotes initial coding skill of Boy $i$.
Next line contains a single integer $Q$ representing the number of queries.
Next $Q$ lines contain queries which can be any of the two types.
$1 \leq N, Q \leq 10^{5} $
$1 \leq A_{i}, v \leq 10^{9}$
$1 \leq l \leq r \leq N$
$1 \leq k \leq r\ -\ l + 1$
For each query of type 2, output a single integer corresponding to the answer in a single line.
5
1 2 3 4 5
3
2 2 4 2
1 3 6
2 2 4 2
3
4
树套树裸题。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
using std::map;
using std::min;
using std::find;
using std::pair;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
#define lc root<<1
#define rc root<<1|1
const int N = 100010;
const int INF = 0x3f3f3f3f;
struct Node {
int v, s, c;
Node *ch[2];
inline void push_up() {
s = ch[0]->s + ch[1]->s + c;
}
inline void setc(int _v, int _s, Node *p) {
v = _v, s = c = _s, ch[0] = ch[1] = p;
}
inline int cmp(int x) const {
if (x == v) return -1;
return x > v;
}
};
int sum, arr[N];
struct SBT {
int top;
Node *null, *tail, *pool[N], stack[N << 5], *ptr[N << 2];
inline void init(int n) {
top = 0;
tail = &stack[0];
null = tail++;
null->setc(0, 0, NULL);
for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
seg_built(1, 1, n);
}
inline Node *newNode(int v) {
Node *p = !top ? tail++ : pool[--top];
p->setc(v, 1, null);
return p;
}
inline void rotate(Node *&x, int d) {
Node *k = x->ch[!d]; x->ch[!d] = k->ch[d]; k->ch[d] = x;
k->s = x->s; x->push_up(); x = k;
}
inline void Maintain(Node *&x, int d) {
if (!x->ch[d]->s) return;
if (x->ch[d]->ch[d]->s > x->ch[!d]->s) rotate(x, !d);
else if (x->ch[d]->ch[!d]->s > x->ch[!d]->s) rotate(x->ch[d], d), rotate(x, !d);
else return;
Maintain(x, 0), Maintain(x, 1);
}
inline void insert(Node* &x, int key) {
if (!x->s) { x = newNode(key); return; }
int d = x->cmp(key);
x->s++;
if (-1 == d) { x->c++; return; }
insert(x->ch[d], key);
x->push_up();
Maintain(x, d);
}
inline void erase(Node* &x, int key){
if (!x->s) return;
int d = x->cmp(key);
x->s--;
if (-1 == d) {
if (x->c > 1) {
x->c--;
} else if (!x->ch[0]->s || !x->ch[1]->s) {
pool[top++] = x;
x = x->ch[0]->s ? x->ch[0] : x->ch[1];
} else {
Node *ret = x->ch[1];
for (; ret->ch[0]->s; ret = ret->ch[0]);
erase(x->ch[1], x->v = ret->v);
}
} else {
erase(x->ch[d], key);
}
if (x->s) x->push_up();
}
inline int sbt_rank(Node *x, int key) {
int t, cur = 0;
for (; x->s;) {
t = x->ch[0]->s;
if (key < x->v) x = x->ch[0];
else if (key >= x->v) cur += x->c + t, x = x->ch[1];
}
return cur;
}
inline void seg_built(int root, int l, int r) {
ptr[root] = null;
for (int i = l; i <= r; i++) insert(ptr[root], arr[i]);
if (l == r) return;
int mid = (l + r) >> 1;
seg_built(lc, l, mid);
seg_built(rc, mid + 1, r);
}
inline void seg_query(int root, int l, int r, int x, int y, int val) {
if (x > r || y < l) return;
if (x <= l && y >= r) {
sum += sbt_rank(ptr[root], val);
return;
}
int mid = (l + r) >> 1;
seg_query(lc, l, mid, x, y, val);
seg_query(rc, mid + 1, r, x, y, val);
}
inline void seg_modify(int root, int l, int r, int pos, int val) {
if (pos > r || pos < l) return;
erase(ptr[root], arr[pos]);
insert(ptr[root], val);
if (l == r) return;
int mid = (l + r) >> 1;
seg_modify(lc, l, mid, pos, val);
seg_modify(rc, mid + 1, r, pos, val);
}
inline void kth(int n, int a, int b, int k) {
int l = 0, r = INF;
while (l < r) {
sum = 0;
int mid = (l + r) >> 1;
seg_query(1, 1, n, a, b, mid);
if (sum < k) l = mid + 1;
else r = mid;
}
printf("%d\n", l);
}
inline void solve(int n) {
init(n);
int m, a, b, c, k;
scanf("%d", &m);
while (m--) {
scanf("%d", &a);
if (2 == a) {
scanf("%d %d %d", &b, &c, &k);
kth(n, b, c, k);
} else {
scanf("%d %d", &b, &c);
seg_modify(1, 1, n, b, c);
arr[b] = c;
}
}
}
}go;
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int n;
while (~scanf("%d", &n)) {
go.solve(n);
}
return 0;
}
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原文地址：http://www.cnblogs.com/GadyPu/p/4796303.html
