D-query SPOJ - DQUERY （莫队算法裸题）

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
Input
5
1 1 2 1 3
3
1 5
2 4
3 5

Output
3
2
3

题意：
给你一个长度为n的数组，和m个询问，对于每一个询问，请你输出数组在 l 到 r 区间中，有多少个不同的数字。
思路：

用一个flag[i] 数组 表示当前区间中出现了多少次i数字。

利用flag数组进行常规的add、del 转移即可。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
pll ans[maxn];
ll Ans = 0ll;
int l = 0;
int r = 0;
struct node {
    int l, r, id;
} a[maxn];
int pos[maxn];
int n, m;
int len;
bool cmp(node aa, node bb)
{
    if (pos[aa.l] == pos[bb.l]) {
        return aa.r < bb.r;
    } else {
        return pos[aa.l] < pos[bb.l];
    }
}
int col[maxn];
int flag[maxn];
void add(int x)
{
    if (flag[col[x]]++ == 0) {
        Ans++;
    }
}
void del(int x)
{
    if (--flag[col[x]] == 0) {
        Ans--;
    }
}
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    gg(n); 
    len = (int)(sqrt(n));
    repd(i, 1, n) {
        gg(col[i]);
    }
    gg(m);
    repd(i, 1, m) {
        gg(a[i].l);
        gg(a[i].r);
        a[i].id = i;
        pos[i] = i / len;
    }
    sort(a + 1, a + 1 + m, cmp);
    repd(i, 1, m) {
        while (l > a[i].l) {
            l--;
            add(l);
        }
        while (r < a[i].r) {
            r++;
            add(r);
        }
        while (l < a[i].l) {
            del(l);
            l++;
        }
        while (r > a[i].r) {
            del(r);
            r--;
        }
        ans[a[i].id].fi = Ans;
    }
    repd(i, 1, m) {
        printf("%lld\n", ans[i].fi);
    }
    return 0;
}

inline void getInt(int *p)
{
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    } else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}