Spoj 3267. D-query (Chairman tree or tree-like array offline)

A-D-queryTime limit:MS Memory Limit:0KB 64bit IO Format:%lld & %llu SubmitStatusPractice Spoj dqueryAppoint Description:System Crawler (2014-12-06)

Description

中文版	Vietnamese

Given a sequence of n numbers a1, A2, ..., an and a number of d-queries. A d-query is a pair (I, J) (1≤i≤j≤n). For each d-query (i, J), you had to return the number of distinct elements in the subsequence ai, ai+1, ..., AJ.

Input

• Line 1:n (1≤n≤30000).
• Line 2:N numbers a1, A2, ..., an (1≤ai≤106).
• Line 3:q (1≤q≤200000), the number of d-queries.
• The next q lines, each line contains 2 numbers I, J representing a D-query (1≤i≤j≤n).

Output

• For each d-query (i, j), print the number of distinct elements in the subsequence ai, ai+1, ..., AJ in a.

Example

Input Output

is to ask the number of different elements in the interval, or you can use a tree-like array to operate offline or use the Chairman tree

Chairman Tree Edition:

/************************************************************************* > File Name:cf.cpp > AUTHOR:ACVCL A > QQ: > Mail: [email protected] > Created time:1949äê10ôâ1èõðçæúò»0ê±0 ö0ãë************************************************************************/#include <iostream> #include <algorithm> #include <cstdio> #include <vector> #include <cstring> #include <map># include<queue> #include <stack> #include <string> #include <cstdlib> #include <ctime># Include<set> #include <math.h>using namespace std;typedef long long ll;const int MAXN = 3e4 + Ten; #define REP (i,a    , b) for (int i= (a); i<= (b); i++) #define PB push_backint tot,n,m,date[maxn],root[maxn];struct chairtnode{int s,rc,lc; Chairtnode (int s=0,int lc=0,int rc=0): s (s), LC (LC), RC (RC) {}};chairtnode chairt[maxn*20];int newnode (int sum,int Lson,    int rson) {int rt=++tot;    Chairt[rt]=chairtnode (Sum,lson,rson); return RT;} void Insert (int &rt,int pre_rt,int pos,int l,int r,int val) {Chairtnode &t=chairT[pre_rt];    Rt=newnode (T.S+VAL,T.LC,T.RC);    if (l==r) return;    int mid= (L+R) >>1;    if (pos<=mid) Insert (chairt[rt].lc,t.lc,pos,l,mid,val); else Insert (chairt[rt].rc,t.rc,pos,mid+1,r,val);}    int Query (int rt,int l,int r,int pos)//query sum of [Pos,r]{if (L==pos) return chairt[rt].s;    int mid= (L+R) >>1;    if (Pos<=mid) return Query (Chairt[rt].lc,l,mid,pos) +chairt[chairt[rt].rc].s; Return Query (Chairt[rt].rc,mid+1,r,pos);}        int main () {while (~SCANF ("%d", &n)) {Rep (i,1,n) {scanf ("%d", date+i);        } tot=root[0]=0;        map<int,int>q;        int t; for (int i=1;i<=n;i++) {if (!q[date[i]) {Insert (root[i],root[i-1],i,1,n,1);//number of positions I plus 1}els            e{Insert (t,root[i-1],q[date[i]],1,n,-1);//has already appeared before, minus the previous insert (root[i],t,i,1,n,1);//Add to the present this time}        Q[date[i]]=i;      } int QL,QR;  scanf ("%d", &m);            while (m--) {scanf ("%d%d", &AMP;QL,&AMP;QR);        printf ("%d\n", Query (ROOT[QR],1,N,QL)); }} return 0;}

Tree-shaped array version:

#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include < algorithm> #include <map>const int maxn1=3e4+5;const int maxn2=2e5+5;typedef Long long ll;using namespace std; int l[maxn2],r[maxn2],loc[maxn1],a[maxn1],id[maxn2],n;int c[maxn1],ans[maxn2];inline int lowbit (int x) {return x& -X;}        void Update (int x,int d) {while (x<=n) {c[x]+=d;    X+=lowbit (x);    }}int Query (int x) {int sum=0;        while (x>0) {sum+=c[x];    X-=lowbit (x); } return sum;} BOOL CMP (int I,int j) {return r[i]<r[j];}        int main () {while (~SCANF ("%d", &n)) {memset (loc,0,sizeof (Loc));        memset (c,0,sizeof (C));        map<int,int>q;        Q.clear ();            for (int i=1;i<=n;i++) {scanf ("%d", a+i);            Update (i,1);            if (!q[a[i]]) {q[a[i]]=i;        }} int m;        scanf ("%d", &m); for (int i=1;i<=m;i++)       {scanf ("%d%d", l+i,r+i);        Id[i]=i;        } sort (id+1,id+1+m,cmp);        /* The main idea is to drag the count as far as possible to the back */int right=1;            for (int i = 1; I <= m; ++i) {int x=id[i]; while (Right<=r[x]) {if (q[a[right]]!=right) {update (Q[A[RIGHT]],-1);/* in the back                Now, let the front count offset, in the back count, to ensure that the number of query interval is not omitted */q[a[right]]=right;            } ++right;            } Right=r[x];        Ans[x]=query (R[x])-query (l[x]-1);    } for (int i=1;i<=m;i++) printf ("%d\n", Ans[i]); } return 0;}

Spoj 3267. D-query (Chairman tree or tree-like array offline)