题目:http://www.spoj.com/problems/DQUERY/en/
题意:给定一个数组,每次询问一个区间内有多少不同的元素
思路:莫队算法裸题
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 30010;
struct node
{
int l, r, id;
}q[N*10];
int n, m, unit, tmp;
int res[N*10], num[N];
int arr[N], brr[N];
void add(int i)
{
if(! num[i]) tmp++;
num[i]++;
}
void del(int i)
{
num[i]--;
if(! num[i]) tmp--;
}
void solve()
{
unit = (int)sqrt(n);
sort(q + 1, q + 1 + m, [](node a, node b){return a.l/unit != b.l/unit ? a.l/unit < b.l/unit : a.r < b.r;});
int l = 1, r = 0;
tmp = 0;
for(int i = 1; i <= m; i++)
{
while(r < q[i].r) add(arr[++r]);
while(r > q[i].r) del(arr[r--]);
while(l < q[i].l) del(arr[l++]);
while(l > q[i].l) add(arr[--l]);
res[q[i].id] = tmp;
}
for(int i = 1; i <= m; i++) printf("%d\n", res[i]);
}
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &arr[i]), brr[i] = arr[i];
sort(brr + 1, brr + 1 + n);
for(int i = 1; i <= n; i++) arr[i] = lower_bound(brr + 1, brr + 1 + n, arr[i]) - brr;
scanf("%d", &m);
for(int i = 1; i <= m; i++) scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
solve();
return 0;
}