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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 5 1 1 2 1 3 3 1 5 2 4 3 5 Output 3 2 3
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|1 const 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; }
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原文地址：http://www.cnblogs.com/dyzll/p/5920839.html
