树套树总结
最近做题发现自己并不知道什么时候该用树套树，就来总结一下
一、静态整体kth
排序输出
sort(a+1,a+n+1);
printf("%d\n",a[k]);
时间复杂度O(nlogn) 空间复杂度O(n)
二、动态整体kth
权值线段树+二分
查询时先查询左子树和sum，比较k和sum的大小：若k<=sum则说明第k小数在左子树中，递归查询左子树；
否则，这个数对应的就是右子树中第k-sum小的数，k-=sum，递归查询右子树。
时间复杂度O(nlogn) 空间复杂度O(n)
应该是这么写的吧……甚至离散化刚开始都写错了
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int read(){
int x = 1,a = 0;char ch = getchar();
while (ch < '0'||ch > '9'){if (ch == '-') x = -1;ch = getchar();}
while (ch >= '0'&&ch <= '9'){a = a*10+ch-'0';ch = getchar();}
return x*a;
}
const int maxn = 1e5+10;
int sum[maxn << 1];
int ls(int x){return x << 1;}
int rs(int x){return x << 1 | 1;}
void modify(int x,int l,int r,int p,int k){
sum[x] += k;
if (l == r) return;
int mid = (l+r >> 1);
if (p <= mid) modify(ls(x),l,mid,p,k);
else modify(rs(x),mid+1,r,p,k);
}
int query(int x,int l,int r,int k){
// cout<<l<<" "<<r<<" "<<sum[ls(x)]<<" "<<k<<endl;
int mid = (l+r >> 1);
if (l == r) return l;
if (sum[ls(x)] >= k) return query(ls(x),l,mid,k);
else return query(rs(x),mid+1,r,k-sum[ls(x)]);
}
int n,m;
int a[maxn],b[maxn];
int main(){
n = read(),m = read();
for (int i = 1;i <= n;i++) a[i] = b[i] = read();
sort(b+1,b+n+1);
int len = unique(b+1,b+n+1)-b-1;
for (int i = 1;i <= n;i++) a[i] = lower_bound(b+1,b+len+1,a[i])-b;
for (int i = 1;i <= n;i++) modify(1,1,len,a[i],1);
for (int i = 1;i <= m;i++){
int op = read();
if (op == 0){
int x = read(),k = read();
modify(1,1,len,a[x],-1);
a[x] = k;
modify(1,1,len,a[x],1);
}
if (op == 1){
int k = read();
printf("%d\n",b[query(1,1,len,k)]);
}
}
return 0;
}
三、静态区间kth
对每个点以其前缀开一棵权值线段树，那么任意一段区间均可以表示成为两棵权值线段树作差，即R位置的线段树减去L-1位置上的线段树
每个点开一棵线段树空间复杂度\(O(n^2)\)，MLE，考虑到后一个位置相比于前一个位置的更改只有logn个节点，所以使用主席树
时间复杂度O(nlogn) 空间复杂度O(nlogn)
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int read(){
int x = 1,a = 0;char ch = getchar();
while (ch < '0'||ch > '9'){if (ch == '-') x = -1;ch = getchar();}
while (ch >= '0'&&ch <= '9'){a = a*10+ch-'0';ch = getchar();}
return x*a;
}
const int maxn = 1e5+10;
int sum[maxn << 1];
int tot,root[maxn];
struct node{
int ls,rs,val;
}tree[maxn*30];
void modify(int &now,int lst,int l,int r,int p,int k){
if (!now) now = ++tot;
tree[now].val = tree[lst].val + k;
if (l == r) return;
int mid = (l+r >> 1);
if (p <= mid) tree[now].rs = tree[lst].rs,modify(tree[now].ls,tree[lst].ls,l,mid,p,k);
else tree[now].ls = tree[lst].ls,modify(tree[now].rs,tree[lst].rs,mid+1,r,p,k);
}
int query(int now,int lst,int l,int r,int k){
if (!now) return 0;
int mid = (l+r >> 1);
if (l == r) return l;
int res = tree[tree[now].ls].val-tree[tree[lst].ls].val;
if (res >= k) return query(tree[now].ls,tree[lst].ls,l,mid,k);
else return query(tree[now].rs,tree[lst].rs,mid+1,r,k-res);
}
int n,m;
int a[maxn],b[maxn];
int main(){
n = read(),m = read();
for (int i = 1;i <= n;i++) a[i] = b[i] = read();
sort(b+1,b+n+1);
int len = unique(b+1,b+n+1)-b-1;
for (int i = 1;i <= n;i++) a[i] = lower_bound(b+1,b+len+1,a[i])-b;
for (int i = 1;i <= n;i++) modify(root[i],root[i-1],1,len,a[i],1);
for (int i = 1;i <= m;i++){
int l = read(),r = read(),k = read();
printf("%d\n",b[query(root[r],root[l-1],1,len,k)]);
}
return 0;
}
四、动态区间kth
还是要想办法维护前缀和。如果只是同3的前缀和的话，就要对前缀和进行O(nlogn)的单次修改，显然TLE。
这里考虑用树状数组维护前缀和。修改时，可以只修改logn个位置，复杂度\(O(log^2n)\)
查询时，依旧是R位置减去L-1位置，这时候不再是两棵线段树作差，而是log棵线段树与log棵线段树作差；跳的时候，log个节点一起跳到左子树/右子树
时间复杂度\(O(nlog^2n)\) 空间复杂度O(nlogn)
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int read(){
int x = 1,a = 0;char ch = getchar();
while (ch < '0'||ch > '9'){if (ch == '-') x = -1;ch = getchar();}
while (ch >= '0'&&ch <= '9'){a = a*10+ch-'0';ch = getchar();}
return x*a;
}
const int maxn = 1e5+10;
int n,m;
int a[maxn],b[maxn];
int ql[maxn],qr[maxn],qk[maxn];
struct node{
int ls,rs,sum;
}tree[maxn*30];
int tot,a1[maxn],a2[maxn],cnt1,cnt2;
void add(int &x,int l,int r,int p,int k){
if (!x) x = ++tot;
tree[x].sum += k;
int mid = (l+r >> 1);
if (l == r) return;
if (p <= mid) add(tree[x].ls,l,mid,p,k);
else add(tree[x].rs,mid+1,r,p,k);
}
int root[maxn],len;
void build(int x,int k){for (int i = x;i <= n;i += i&-i) add(root[i],1,len,a[x],k);}
int getkth(int l,int r,int k){
int res = 0,mid = (l+r >> 1);
if (l == r) return l;
for (int i = 1;i <= cnt1;i++) res += tree[tree[a1[i]].ls].sum;
for (int i = 1;i <= cnt2;i++) res -= tree[tree[a2[i]].ls].sum;
if (res >= k){
for (int i = 1;i <= cnt1;i++) a1[i] = tree[a1[i]].ls;
for (int i = 1;i <= cnt2;i++) a2[i] = tree[a2[i]].ls;
return getkth(l,mid,k);
}
else{
for (int i = 1;i <= cnt1;i++) a1[i] = tree[a1[i]].rs;
for (int i = 1;i <= cnt2;i++) a2[i] = tree[a2[i]].rs;
return getkth(mid+1,r,k-res);
}
}
int query(int l,int r,int k){
cnt1 = 0,cnt2 = 0;
for (int i = r;i;i -= i&-i) a1[++cnt1] = root[i];
for (int i = l-1;i;i -= i&-i) a2[++cnt2] = root[i];
return getkth(1,len,k);
}
int cnt;
int main(){
n = read(),m = read();
for (int i = 1;i <= n;i++) a[i] = b[++cnt] = read();
for (int i = 1;i <= m;i++){
char op[10];scanf ("%s",op);
if (op[0] == 'Q') ql[i] = read(),qr[i] = read(),qk[i] = read();
if (op[0] == 'C') ql[i] = read(),qk[i] = b[++cnt] = read();
}
sort(b+1,b+cnt+1);
len = unique(b+1,b+cnt+1)-b-1;
for (int i = 1;i <= n;i++) a[i] = lower_bound(b+1,b+len+1,a[i])-b;
for (int i = 1;i <= n;i++) if (!qr[i]) qk[i] = lower_bound(b+1,b+len+1,qk[i])-b;
for (int i = 1;i <= n;i++) build(i,1);
for (int i = 1;i <= m;i++){
if (qr[i]) printf("%d\n",b[query(ql[i],qr[i],qk[i])]);
else build(ql[i],-1),a[ql[i]] = qk[i],build(ql[i],1);
}
return 0;
}
