Luogu P1903 [国家集训队]数颜色 or 维护队列

标准的带修莫队。。。咕到了现在$qwq$

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莫队是对询问排序来优化复杂度的（不带修就是对询问区间$[l,r]$排序）。。

那么现在带修了，我们再可以维护一个时间维度$tm$：对于每个询问，每次回答前先检查时间指针是否与询问的时间对应，不对应则按操作时间修改。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<queue>
#include<map>
#include<set>
#define ull unsigned long long
#define ll long long
#define R register int
#define pause (for(R i=1;i<=10000000000;++i))
#define OUT freopen("out.out","w",stdout);
using namespace std;
namespace Fread {
    //static char B[1<<15],*S=B,*D=B;
    //#define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
    inline int g() {
        R ret=,fix=; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-:fix;
        do ret=ret*+(ch^); while(isdigit(ch=getchar())); return ret*fix;
    } inline bool isempty(const char& ch) {return ch<=||ch>=;}
    inline void gs(char* s) {register char ch; while(isempty(ch=getchar())); do *s++=ch; while(!isempty(ch=getchar()));}
}using Fread::g; using Fread::gs;
const int M=;
int n,m,T,tim,cl[M*],cnt,cntq,a[M],crt[M],pos[M],ans[M],anss,l=,r=;
struct node { int l,r,tm,rk;
    inline bool operator <(const node& that) {
        return pos[l]==pos[that.l]?(pos[r]==pos[that.r]?tm<that.tm:(pos[l]&)?r<that.r:r>that.r):l<that.l;
        //return pos[l]==pos[that.l]?(pos[r]==pos[that.r]?tm<that.tm:r<that.r):l<that.l;
    }
    #define l(i) s[i].l
    #define r(i) s[i].r
    #define tm(i) s[i].tm
    #define rk(i) s[i].rk
}s[M];
struct que{int pos,nw,od;}q[M];
inline void change(int pos,int inc) {inc>?anss+=(++cl[pos]==):anss-=(--cl[pos]==);}
inline void upd(int pos,int nw) {
    if(pos>=l&&pos<=r) change(a[pos],-),change(nw,); a[pos]=nw;
}
signed main() {
#ifdef JACK
    freopen("NOIPAK++.in","r",stdin);
#endif
    n=g(),m=g(); T=pow(n,2.0/);
    for(R i=;i<=n;++i) a[i]=g(); memcpy(crt,a,sizeof(int)*(n+));
    for(R i=;i<=n;++i) pos[i]=(i-)/T+;
    for(R i=;i<=m;++i) { register char ch; while(!isalpha(ch=getchar()));
        R l=g(),r=g(); if(ch=='Q') l(++cnt)=l,r(cnt)=r,tm(cnt)=cntq,rk(cnt)=cnt;
        else q[++cntq].pos=l,q[cntq].nw=r,q[cntq].od=crt[l],crt[l]=r;
    } sort(s+,s+cnt+); for(R i=;i<=m;++i) {
        while(tim<tm(i)) ++tim,upd(q[tim].pos,q[tim].nw);
        while(tim>tm(i)) upd(q[tim].pos,q[tim].od),--tim;
        while(l<l(i)) change(a[l],-),++l;
        while(l>l(i)) change(a[--l],);
        while(r<r(i)) change(a[++r],);
        while(r>r(i)) change(a[r],-),--r;
        ans[rk(i)]=anss;
    } for(R i=;i<=cnt;++i) printf("%d\n",ans[i]);
}

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2019.07.03