SP3267 DQUERY - D-query(离线树状数组)
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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;
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