SPOJ 3267 求区间不同数的个数

题意：给定一个数列，每次查询一个区间不同数的个数。

做法：离线+BIT维护。将查询按右端点排序。从左到右扫，如果该数之前出现过，则将之前出现过的位置相应删除；当前位置则添加1。这样做就保证每次扫描到的某一位置，以该位置为右端点的区间都相应地确定了。

/*
 *Author:       Zhaofa Fang
 *Created time: 2013-08-25-22.29 星期日
 */
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;

typedef long long ll;
typedef pair<int,int> PII;
#define DEBUG(x) cout<< #x << ':' << x << endl
#define FOR(i,s,t) for(int i = (s);i <= (t);i++)
#define FORD(i,s,t) for(int i = (s);i >= (t);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define REPD(i,n) for(int i=(n-1);i>=0;i--)
#define PII pair<int,int>
#define PB push_back
#define ft first
#define sd second
#define lowbit(x) (x&(-x))
#define INF (1<<30)
#define eps (1e-8)

const int maxq = 200011;
const int maxn = 30011;
int a[maxn],C[maxn],last[1000011];
int ans[maxq];
void init(){
    memset(C,0,sizeof(C));
    memset(last,-1,sizeof(last));
}
struct Query{
    int l,r;
    int idx;
    bool operator < (const Query & rhs)const{
        return r < rhs.r;
    }
}Q[maxq];

void add(int x,int val){
    while(x<maxn){
        C[x] += val;
        x += lowbit(x);
    }
}
int sum(int x){
    int res = 0;
    while(x > 0){
        res += C[x];
        x -= lowbit(x);
    }
    return res;
}
int main(){
    //freopen("in","r",stdin);
    //freopen("out","w",stdout);
    int n;
    while(~scanf("%d",&n)){
        init();
        FOR(i,1,n)scanf("%d",&a[i]);
        int q;
        scanf("%d",&q);
        REP(i,q){
            scanf("%d%d",&Q[i].l,&Q[i].r);
            Q[i].idx = i;
        }
        sort(Q,Q+q);
        int pre = 1;
        REP(i,q){
            FOR(j,pre,Q[i].r){
                if(last[a[j]]==-1){
                    add(j,1);
                }else {
                    add(last[a[j]],-1);
                    add(j,1);
                }
                last[a[j]] = j;
            }
            ans[Q[i].idx] = sum(Q[i].r)-sum(Q[i].l-1);
            pre = Q[i].r+1;
        }
        REP(i,q)printf("%d
",ans[i]);
    }
    return 0;
}