莫队算法 Gym - 100496D Data Mining

 

题目传送门

  1 /*
  2     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
  3     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:)
  4 */
  5 #include <cstdio>
  6 #include <algorithm>
  7 #include <map>
  8 #include <set>
  9 #include <vector>
 10 #include <cmath>
 11 #include <cstring>
 12 using namespace std;
 13 
 14 const int MAXN = 2e6 + 10;
 15 const int INF = 0x3f3f3f3f;
 16 map<int, int> table;
 17 vector<int> V[MAXN];
 18 int cnt[MAXN];
 19 int a[MAXN];
 20 int ans[MAXN];
 21 struct Data
 22 {
 23     int b, l, r, id;
 24     Data () {}
 25     Data (int b, int l, int r, int id) : b (b), l (l), r (r), id (id) {};
 26 }data[MAXN];
 27 int n, m;
 28 int ret;
 29 
 30 bool cmp(Data x, Data y)
 31 {
 32     if (x.b == y.b)    return x.r < y.r;
 33     else    return x.b < y.b;
 34 }
 35 
 36 void updata(int v)
 37 {
 38     if (v == 1)    ret++;
 39     else if (v == 0)    ret--;
 40 }
 41 
 42 void Modui(void)
 43 {
 44     sort (data+1, data+1+m, cmp);
 45     memset (cnt, 0, sizeof (cnt));
 46 
 47     int l = 1, r = 0;    ret = 0;
 48     for (int i=1; i<=m; ++i)
 49     {
 50         while (data[i].l < l)
 51         {
 52             ++cnt[a[--l]];
 53             if (cnt[a[l]] == 1)    ret++;
 54         }
 55         while (data[i].l > l)
 56         {
 57             --cnt[a[l]];
 58             if (cnt[a[l]] == 0)    ret--;
 59             l++;
 60         }
 61         while (data[i].r > r)
 62         {
 63             ++cnt[a[++r]];
 64             if (cnt[a[r]] == 1)    ret++;
 65         }
 66         while (data[i].r < r)
 67         {
 68             --cnt[a[r]];
 69             if (cnt[a[r]] == 0)    ret--;
 70             r--;
 71         }
 72 
 73         ans[data[i].id] = ret;
 74     }
 75 
 76     for (int i=1; i<=m; ++i)
 77     {
 78         printf ("%d\n", ans[i]);
 79     }
 80 }
 81 
 82 int main(void)        //Gym - 100496D Data Mining
 83 {
 84     // freopen ("D.in", "r", stdin);
 85     freopen ("data.in", "r", stdin);
 86     freopen ("data.out", "w", stdout);
 87 
 88     while (scanf ("%d", &n) == 1)
 89     {
 90         table.clear ();
 91         for (int i=1; i<=n; ++i)    V[i].clear ();
 92 
 93         int num = 0;
 94         for (int i=1; i<=n; ++i)
 95         {
 96             scanf ("%d", &a[i]);
 97             a[i] = table[a[i]] ? table[a[i]] : table[a[i]] = ++num;
 98             V[a[i]].push_back (i);
 99         }
100 
101         int block = (int) sqrt (n * 1.0);
102         scanf ("%d", &m);
103         for (int i=1; i<=m; ++i)
104         {
105             int l, r;    scanf ("%d%d", &l, &r);
106             r = l + r - 1;
107             int pos = lower_bound (V[a[r]].begin (), V[a[r]].end (), l) - V[a[r]].begin ();
108             r = V[a[r]][pos];
109             data[i] = Data (l / block, l, r, i);
110         }
111 
112         Modui ();
113     }
114 
115     return 0;
116 }

 

posted @ 2015-07-14 10:19  Running_Time  阅读(312)  评论(0)  编辑  收藏  举报