莫队算法 Gym - 100496D Data Mining

题目传送门

 /*
     题意：从i开始，之前出现过的就是之前的值，否则递增，问第p个数字是多少
     莫队算法：先把a[i+p-1]等效到最前方没有它的a[j]，问题转变为求[l, r]上不重复数字有几个，裸莫队:)
 */
 #include <cstdio>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <vector>
 #include <cmath>
 #include <cstring>
 using namespace std;

 const int MAXN = 2e6 + ;
 const int INF = 0x3f3f3f3f;
 map<int, int> table;
 vector<int> V[MAXN];
 int cnt[MAXN];
 int a[MAXN];
 int ans[MAXN];
 struct Data
 {
     int b, l, r, id;
     Data () {}
     Data (int b, int l, int r, int id) : b (b), l (l), r (r), id (id) {};
 }data[MAXN];
 int n, m;
 int ret;

 bool cmp(Data x, Data y)
 {
     if (x.b == y.b)    return x.r < y.r;
     else    return x.b < y.b;
 }

 void updata(int v)
 {
     if (v == )    ret++;
     else if (v == )    ret--;
 }

 void Modui(void)
 {
     sort (data+, data++m, cmp);
     memset (cnt, , sizeof (cnt));

     int l = , r = ;    ret = ;
     for (int i=; i<=m; ++i)
     {
         while (data[i].l < l)
         {
             ++cnt[a[--l]];
             if (cnt[a[l]] == )    ret++;
         }
         while (data[i].l > l)
         {
             --cnt[a[l]];
             if (cnt[a[l]] == )    ret--;
             l++;
         }
         while (data[i].r > r)
         {
             ++cnt[a[++r]];
             if (cnt[a[r]] == )    ret++;
         }
         while (data[i].r < r)
         {
             --cnt[a[r]];
             if (cnt[a[r]] == )    ret--;
             r--;
         }

         ans[data[i].id] = ret;
     }

     for (int i=; i<=m; ++i)
     {
         printf ("%d\n", ans[i]);
     }
 }

 int main(void)        //Gym - 100496D Data Mining
 {
     // freopen ("D.in", "r", stdin);
     freopen ("data.in", "r", stdin);
     freopen ("data.out", "w", stdout);

     while (scanf ("%d", &n) == )
     {
         table.clear ();
         for (int i=; i<=n; ++i)    V[i].clear ();

         int num = ;
         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &a[i]);
             a[i] = table[a[i]] ? table[a[i]] : table[a[i]] = ++num;
             V[a[i]].push_back (i);
         }

         int block = (int) sqrt (n * 1.0);
         scanf ("%d", &m);
         for (int i=; i<=m; ++i)
         {
             int l, r;    scanf ("%d%d", &l, &r);
             r = l + r - ;
             int pos = lower_bound (V[a[r]].begin (), V[a[r]].end (), l) - V[a[r]].begin ();
             r = V[a[r]][pos];
             data[i] = Data (l / block, l, r, i);
         }

         Modui ();
     }

     return ;
 }