SP3267 DQUERY - D-query（离线树状数组）

给你一个序列，询问一个区间内有多少个不同的数字。

经典离线树状数组，类似于HH（憨憨）的项链，把询问按照右端点排序，保证在每个数字最后一次出现的位置上加1

如果有询问的右端点等于当前加到的数字下标，就对它求一次和并累加进答案，直到所有的查询都被处理。

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=1000005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read()
{
    char ch=getchar();int s=0,w=1;
    while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}
    while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
    return s*w;
}
inline void write(int x)
{
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);
}

int n,a[maxn],q,ans[maxn],pre[maxn];
struct node{
    int l,r,id;
    bool operator < (const node& o){
        if(r!=o.r) return r<o.r;
        else return l<o.l;
    }
}ask[maxn];

int tr[maxn*4];
int sum(int x){
    int res=0;
    for(;x>0;x-=lowbit(x)) res+=tr[x];
    return res;
}
void add(int x,int y){
    for(;x<=1000000;x+=lowbit(x)) tr[x]+=y;
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    sc(n);
    re(i,1,n) sc(a[i]);
    sc(q);
    re(i,1,q) scc(ask[i].l,ask[i].r),ask[i].id=i;
    sort(ask+1,ask+1+q);
    int k=1;
    re(i,1,n){
        if(pre[a[i]]!=0) add(pre[a[i]],-1);
        add(i,1);
        pre[a[i]]=i;
        while(ask[k].r==i){
            ans[ask[k].id]=sum(ask[k].r)-sum(ask[k].l-1);
            k++;
            if(k>q) break;
        }
    }
    re(i,1,q) pr(ans[i]),prn();
    return 0;
}

标签：typedef,return,int,SP3267,离线,DQUERY,const,lld,define
来源： https://www.cnblogs.com/oneman233/p/11488393.html