luogu 1972 小H的项链 莫队

1.莫队算法 TLE 80

#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i=x;i<=y;i++)
using namespace std;
const int N=,M=;
int h[N],n,m,a[N],ans[M];
struct node{
    int l,r,id;
    bool operator<(const node&b)const{
       if(h[l]==h[b.l])return r<b.r;
       return l<b.l;}}q[M];
inline int read(){
    int x=,f=;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
    while(isdigit(ch)){x=(x<<)+(x<<)+(ch^);ch=getchar();}
    return x*f;}
void block(){
    int k=sqrt(n);
    rep(i,,n) h[i]=(i-)/k+;}
int L,R,now,vis[];
//vis实质上记录的是出现次数
void revise(int i,int d){
     if(d==){
         if(vis[a[i]]==)now++;vis[a[i]]++;}
     else{
        vis[a[i]]--;if(!vis[a[i]])now--;}}
int main(){
     n=read();
     rep(i,,n) a[i]=read();
     block();m=read();
     rep(i,,m) q[i].l=read(),q[i].r=read(),q[i].id=i;
     sort(q+,q++m);
     rep(i,,m){
         while(L<q[i].l) revise(L++,-);
         while(L>q[i].l) revise(--L,);
         while(R<q[i].r) revise(++R,);
         while (R>q[i].r) revise(R--,-);
         ans[q[i].id]=now;}
     //注意编号问题，会影响值(unknown reasons)
     rep(i,,m) printf("%d\n",ans[i]);
     return ;
}

2.离线树状数组

别人的代码，没看懂，链接：https://www.cnblogs.com/five20/p/7603849.html

 #include<bits/stdc++.h>
  using namespace std;
  int a[],s[],has[],last[],n,m;
  struct o{
  int x,y,num;
  }q[];
  inline int gi()
  {
      int a=;char x=getchar();bool f=;
     while((x>''||x<'')&&x!='-')x=getchar();
     if(x=='-')x=getchar(),f=;
     while(x>=''&&x<='')a=a*+x-'',x=getchar();
     return f?-a:a;
 }
 bool cmp(o a,o b)
 {return a.y<b.y;}
 inline void add(int k,int a)
 {
     while(k<=n)
     {
         s[k]+=a;
         k+=k&-k;
     }
 }
 int ans(int x)
 {
     int sum=;
     while(x)
     {
         sum+=s[x];
         x-=x&-x;
     }
     return sum;
 }
 int main()
 {
     n=gi();
     for(int i=;i<=n;i++)
     {int A=gi();has[i]=last[A]+;last[A]=i;}
     m=gi();
     for(int i=;i<=m;i++)
     {
         q[i].x=gi();q[i].y=gi();q[i].num=i;
     }
     sort(q+,q+m+,cmp);
     int now=;
     for(int i=;i<=m;i++)
     {
         while(now<=q[i].y)
         {
             now++;
             add(has[now-],);
             add(now,-);
         }
         a[q[i].num]=ans(q[i].x);
     }
     for(int i=;i<=m;i++)
     printf("%d\n",a[i]);
     return ;
 }