SPOJ DQUERY D-query 莫队算法

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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;}

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