数据结构模版题【连这么神的题都沦为模版题了Orz
对数离散化后树状数组套权值线段树。
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define clr(x, c) memset(x, c, sizeof(x))
#define lowbit(x) (x&-x)
#define maxn 10009
#define maxm 20009
#define inf 0x7fffffff
#define k(x) Key[x]
#define t(x) Tree[x]
using namespace std;
inline int read()
{
int x=0, f=1; char ch=getchar();
while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while (isdigit(ch)) x=x*10+ch-'0', ch=getchar();
return x*f;
}
struct node{node *l, *r; int sum;} *blank=new(node), *Tree[maxn], *r1[maxn], *r2[maxn];
struct node2{int v, n;} num[maxm];
bool cmp(node2 a, node2 b){return a.v<b.v;}
int n, m, V, q[maxn][4], Key[maxm], ln, n1, n2, c[maxm];
char ch[5];
void Build(int l, int r, node*&t)
{
if (t==blank) t=new(node), t->l=t->r=blank, t->sum=0;
if (l==r) return;
int mid=(l+r)>>1;
Build(l, mid, t->l), Build(mid+1, r, t->r);
}
void Add(int k, int y, int l, int r, node *u, node*&t)
{
if (t==blank) t=new(node), t->l=t->r=blank, t->sum=0;
t->sum=u->sum+y;
if (l==r) return; int mid=(l+r)>>1;
if (k<=mid)
t->r=u->r, Add(k, y, l, mid, u->l, t->l);
else
t->l=u->l, Add(k, y, mid+1, r, u->r, t->r);
}
inline void Change(int t, int k)
{
node *p; int v=k(t); k(t)=k;
for(int x=t; x<=n; x+=lowbit(x))
Add(v, -1, 1, ln, t(x), p=blank), t(x)=p;
for(int x=t; x<=n; x+=lowbit(x))
Add(k, 1, 1, ln, t(x), p=blank), t(x)=p;
}
inline int Query(int l, int r, int k)
{
n1=n2=0; k--;
for(int x=l-1; x; x-=lowbit(x)) r1[++n1]=t(x);
for(int x=r; x; x-=lowbit(x)) r2[++n2]=t(x);
int L=1, R=ln;
while (L<R)
{
int sum=0, mid=(L+R)>>1;
rep(j, 1, n1) sum-=r1[j]->l->sum;
rep(j, 1, n2) sum+=r2[j]->l->sum;
if (sum<=k)
{
L=mid+1, k-=sum;
rep(j, 1, n1) r1[j]=r1[j]->r;
rep(j, 1, n2) r2[j]=r2[j]->r;
}
else
{
R=mid;
rep(j, 1, n1) r1[j]=r1[j]->l;
rep(j, 1, n2) r2[j]=r2[j]->l;
}
}
return c[L];
}
void Init(){blank->l=blank->r=blank; blank->sum=0;}
int main()
{
n=V=read(), m=read(); Init();
rep(i, 1, n) num[i].v=read(), num[i].n=i;
rep(i, 1, m)
{
scanf("%s", ch);
if (ch[0]=='Q')
q[i][0]=1, q[i][1]=read(), q[i][2]=read(), q[i][3]=read();
else
{
q[i][0]=0, q[i][1]=read(), q[i][2]=read();
++V, num[V].v=q[i][2], num[V].n=n+i;
}
}
sort(num+1, num+V+1, cmp);
c[++ln]=num[1].v, k(num[1].n)=1;
rep(i, 2, V)
{
if (num[i].v != num[i-1].v) c[++ln]=num[i].v;
k(num[i].n)=ln;
}
Build(1, ln, t(0)=blank);
rep(i, 1, n)
{
node *p=t(0);
rep(j, i-lowbit(i)+1, i)
Add(k(j), 1, 1, ln, p, t(i)=blank), p=t(i);
}
rep(i, 1, m)
if (!q[i][0]) Change(q[i][1], k(n+i));
else printf("%d
", Query(q[i][1], q[i][2], q[i][3]));
return 0;
}