一种可以在区间上找名次的数据结构。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
namespace Treap {
#define ls ch[id][0]
#define rs ch[id][1]
const int INF = 2147483647;
const int N = 5e4 + 5;
const int MAXN = 25 * N;
//每个元素会被覆盖若干次线段树有log层，每层的平衡树的长度和都是n，按道理说是精确的nlogn，但这里还是开尽可能大
int ch[MAXN][2], dat[MAXN];
int val[MAXN];
int cnt[MAXN];
int siz[MAXN];
int tot;
inline void Init() {
tot = 0;
}
inline int NewNode(int v, int num) {
int id = ++tot;
ls = rs = 0;
dat[id] = rand();
val[id] = v;
cnt[id] = num;
siz[id] = num;
return id;
}
inline void PushUp(int id) {
siz[id] = siz[ls] + siz[rs] + cnt[id];
}
inline void Rotate(int &id, int d) {
int temp = ch[id][d ^ 1];
ch[id][d ^ 1] = ch[temp][d];
ch[temp][d] = id;
id = temp;
PushUp(ch[id][d]);
PushUp(id);
}
//插入num个v
inline void Insert(int &id, int v, int num) {
if(!id)
id = NewNode(v, num);
else {
if(v == val[id])
cnt[id] += num;
else {
int d = val[id] > v ? 0 : 1;
Insert(ch[id][d], v, num);
if(dat[id] < dat[ch[id][d]])
Rotate(id, d ^ 1);
}
PushUp(id);
}
}
//删除至多num个v
void Remove(int &id, int v, int num) {
if(!id)
return;
else {
if(v == val[id]) {
if(cnt[id] > num) {
cnt[id] -= num;
PushUp(id);
} else if(ls || rs) {
if(!rs || dat[ls] > dat[rs])
Rotate(id, 1), Remove(rs, v, num);
else
Rotate(id, 0), Remove(ls, v, num);
PushUp(id);
} else
id = 0;
} else {
val[id] > v ? Remove(ls, v, num) : Remove(rs, v, num);
PushUp(id);
}
}
}
//查询严格<v的数的个数（和普通平衡树不一样
int GetRank(int id, int v) {
int res = 0;
while(id) {
if(val[id] > v)
id = ls;
else if(val[id] == v) {
res += siz[ls];
break;
} else {
res += siz[ls] + cnt[id];
id = rs;
}
}
return res;
}
//查询树中的<=x的数的个数有至少rk个的最小的x
int GetValue(int id, int rk) {
int res = INF;
while(id) {
if(siz[ls] >= rk)
id = ls;
else if(siz[ls] + cnt[id] >= rk) {
res = val[id];
break;
} else {
rk -= siz[ls] + cnt[id];
id = rs;
}
}
return res;
}
//查询v的前驱的值（<v的第一个节点的值），不存在前驱返回负无穷
int GetPrev(int id, int v) {
int res = -INF;
while(id) {
if(val[id] < v)
res = val[id], id = rs;
else
id = ls;
}
return res;
}
//查询v的后继的值（>v的第一个节点的值），不存在后继返回无穷
int GetNext(int id, int v) {
int res = INF;
while(id) {
if(val[id] > v)
res = val[id], id = ls;
else
id = rs;
}
return res;
}
#undef ls
#undef rs
}
namespace SegmentTree {
#define ls (p<<1)
#define rs (p<<1|1)
const int INF = 2147483647;
const int MAXN = 5e4 + 5;
int n, m, a[MAXN];
struct SegmentTreeNode {
int l, r;
int root;
} st[MAXN * 4];
//线段树的大小是精确的4倍
void Build(int p, int l, int r) {
st[p].l = l, st[p].r = r;
for (int i = l; i <= r ; ++i)
Treap::Insert(st[p].root, a[i], 1);
if(l == r)
return;
int mid = l + r >> 1;
Build(ls, l, mid);
Build(rs, mid + 1, r);
}
void Update(int p, int pos, int v) {
Treap::Remove(st[p].root, a[pos], 1);
Treap::Insert(st[p].root, v, 1);
if (st[p].l == st[p].r)
return;
int mid = st[p].l + st[p].r >> 1;
if(pos <= mid)
Update(ls, pos, v);
else
Update(rs, pos, v);
}
//查询严格<v的数的个数（和普通平衡树不一样
int GetRank(int p, int l, int r, int v) {
if (st[p].l > r || st[p].r < l)
return 0;
if (st[p].l >= l && st[p].r <= r)
return Treap::GetRank(st[p].root, v);
else
//很明显是满足结合律的
return GetRank(ls, l, r, v) + GetRank(rs, l, r, v) ;
}
//查询区间中的<=x的数的个数有至少rk个的最小的x
int GetValue(int l, int r, int k) {
int L = 0, R = 1e8;
while(1) {
int mid = L + R >> 1;
if(L == mid)
return L;
//小于k，也有可能mid是刚刚好的，比如mid有连续的一段
if(GetRank(1, l, r, mid) < k)
L = mid;
else
R = mid;
}
}
int GetPrev(int p, int l, int r, int k) {
if (st[p].l > r || st[p].r < l)
return -INF;
if (st[p].l >= l && st[p].r <= r)
return Treap::GetPrev(st[p].root, k);
else
//很明显是满足结合律的
return max(GetPrev(ls, l, r, k), GetPrev(rs, l, r, k));
}
int GetNext(int p, int l, int r, int k) {
if (st[p].l > r || st[p].r < l)
return INF;
if (st[p].l >= l && st[p].r <= r)
return Treap::GetNext(st[p].root, k);
else
//很明显是满足结合律的
return min(GetNext(ls, l, r, k), GetNext(rs, l, r, k));
}
#undef ls
#undef rs
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
SegmentTree::n = n;
for (int i = 1; i <= n ; ++i)
scanf("%d", &SegmentTree::a[i]);
SegmentTree::Build(1, 1, n);
for (int i = 1; i <= m; ++i) {
int opt, l, k, r, pos;
scanf("%d", &opt);
switch(opt) {
case 1:
scanf("%d%d%d", &l, &r, &k);
printf("%d\n", SegmentTree::GetRank(1, l, r, k) + 1);
break;
case 2:
scanf("%d%d%d", &l, &r, &k);
printf("%d\n", SegmentTree::GetValue(l, r, k));
break;
case 3:
scanf("%d%d", &pos, &k);
SegmentTree::Update(1, pos, k);
SegmentTree::a[pos] = k;
break;
case 4:
scanf("%d%d%d", &l, &r, &k);
printf("%d\n", SegmentTree::GetPrev(1, l, r, k));
break;
case 5:
scanf("%d%d%d", &l, &r, &k);
printf("%d\n", SegmentTree::GetNext(1, l, r, k));
break;
}
}
return 0;
}