题解: 根据题面公式将题目转化成二维平面问题 可持久话不是真的可持久化 只需要将以前修改的点一并放在二维平面中即可 然后问题转化为求二维平面斜正方形里面包含的节点数 然而斜正方形往下搜的复杂度会退化到n^2 所以我们考虑将坐标系旋转一下 然后查询正方形内的节点个数 (挂价函数写挂了 一直WA 气气~~~
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=1e5+10;
const double eps=1e-8;
const int inf=1e9;
#define ll long long
using namespace std;
struct edge{int t,v;edge*next;}e[MAXN<<1],*h[MAXN],*o=e;
void add(int x,int y,int vul){o->t=y;o->v=vul;o->next=h[x];h[x]=o++;}
ll read(){
ll 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;
}
int rt,d,n,q;
typedef struct node{
int p[2],maxx[2],minn[2],c[2],sz;
friend bool operator<(node aa,node bb){
if(aa.p[d]!=bb.p[d])return aa.p[d]<bb.p[d];
return aa.p[d^1]<bb.p[d^1];
}
}node;
node a[MAXN];int b[MAXN];
void up(int x){
int l=a[x].c[0];int r=a[x].c[1];
for(int i=0;i<=1;i++)a[x].minn[i]=a[x].maxx[i]=a[x].p[i];
for(int i=0;i<=1;i++)a[x].minn[i]=min(a[x].minn[i],a[l].minn[i]),a[x].maxx[i]=max(a[x].maxx[i],a[l].maxx[i]);
for(int i=0;i<=1;i++)a[x].minn[i]=min(a[x].minn[i],a[r].minn[i]),a[x].maxx[i]=max(a[x].maxx[i],a[r].maxx[i]);
a[x].sz=a[a[x].c[0]].sz+a[a[x].c[1]].sz+1;
}
int built(int l,int r,int now){
int mid=(l+r)>>1;
d=now;nth_element(a+l,a+mid,a+r+1);
for(int i=0;i<=1;i++)a[mid].minn[i]=a[mid].maxx[i]=a[mid].p[i];
a[mid].c[0]=a[mid].c[1]=0;a[mid].sz=0;
if(l<mid)a[mid].c[0]=built(l,mid-1,now^1);
if(r>mid)a[mid].c[1]=built(mid+1,r,now^1);
up(mid);
return mid;
}
void insert(int &k , int x)
{
if(!k) k = x;
else if(a[x] < a[k]) d ^= 1 , insert(a[k].c[0] , x);
else d ^= 1 , insert(a[k].c[1] , x);
up(k);
}
int ans;
void querty(int x,int x1,int y1,int x2,int y2){
if(!x)return ;
if(x2<a[x].minn[0]||x1>a[x].maxx[0]||y2<a[x].minn[1]||y1>a[x].maxx[1])return ;
if(x1<=a[x].minn[0]&&x2>=a[x].maxx[0]&&y1<=a[x].minn[1]&&y2>=a[x].maxx[1]){ans+=a[x].sz;return ;}
if(a[x].p[0]>=x1&&a[x].p[0]<=x2&&a[x].p[1]>=y1&&a[x].p[1]<=y2){ans++;}
querty(a[x].c[0],x1,y1,x2,y2);
querty(a[x].c[1],x1,y1,x2,y2);
}
char str[11];
int main(){
a[0].minn[0]=a[0].minn[1]=inf;a[0].maxx[0]=a[0].maxx[1]=-inf;
n=read();q=read();int tot=n;
int t,x,k;
for(int i=1;i<=n;i++){b[i]=t=read();a[i].p[0]=i-t;a[i].p[1]=i+t;}
rt=built(1,n,0);
while(q--){
scanf("%s",str);x=read();k=read();
if(str[0]=='Q'){ans=0;querty(rt,x-k-b[x],x-k+b[x],x+k-b[x],x+k+b[x]);printf("%d
",ans);}
else{
tot++;a[tot].c[0]=a[tot].c[1]=0;a[tot].sz=0;
a[tot].p[0]=a[tot].minn[0]=a[tot].maxx[0]=x-k;
a[tot].p[1]=a[tot].minn[1]=a[tot].maxx[1]=x+k;
b[x]=k;
insert(rt,tot);
//if(tot%10000==0)built(1,tot,0);
}
}
return 0;
}