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SPOJ DQUERY - D-query
DQUERY - D-query
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 have 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.
- In 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 single line.
Example
Input51 1 2 1 331 52 43 5Output323
分析：离线树状数组，在线主席树（待写）；
代码：
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 1000000007#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=1e6+10;using namespace std;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}inline ll read(){ ll x=0;int f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f;}int n,m,k,t,pos[maxn],a[maxn],b[maxn],ans[maxn];vector<pii >query[maxn];void add(int x,int y){ for(int i=x;i<=n;i+=(i&(-i))) a[i]+=y;}int get(int x){ int ret=0; for(int i=x;i;i-=(i&(-i))) ret+=a[i]; return ret;}int main(){ int i,j; scanf("%d",&n); rep(i,1,n)scanf("%d",&b[i]); scanf("%d",&m); rep(i,1,m) { int c,d; scanf("%d%d",&c,&d); query[d].pb(mp(c,i)); } rep(i,1,n) { if(pos[b[i]])add(pos[b[i]],-1); pos[b[i]]=i; add(pos[b[i]],1); for(pii x:query[i])ans[x.se]=get(i)-get(x.fi-1); } rep(i,1,m)printf("%d\n",ans[i]); //system("Pause"); return 0;}
SPOJ DQUERY - D-query
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