莫队算法

Beautiful Girl

　　题意

　　给定一个长度为 n 的序列 a[1], a[2], ..., a[n] .

　　m 组询问 (l, r, K) , 求区间 [l, r] 去除重复的数之后的第 K 小.

　　n, m <= 100000 .

　　分析

　　莫队算法 + 值域分块.

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cctype>
 5 #include <cmath>
 6 #include <algorithm>
 7 using namespace std;
 8 #define F(i, a, b) for (register int i = (a); i <= (b); i++)
 9 inline int rd(void) {
10     int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -1;
11     int x = 0; for (; isdigit(c); c = getchar()) x = x*10+c-‘0‘; return x*f;
12 }
13
14 const int N = 100005;
15
16 int n, a[N], Hash[N], tot;
17
18 int m, c, ans[N];
19 struct Qry {
20     int l, r, K, id;
21     friend inline bool operator < (Qry A, Qry B) {
22         int ownA = (A.l+c-1)/c, ownB = (B.l+c-1)/c;
23         return ownA != ownB ? ownA < ownB : A.r < B.r;
24     }
25 }qs[N];
26 int cnt[N], siz[N];
27
28 inline void Add(int x) {
29     x = a[x];
30     cnt[x]++;
31     if (cnt[x] == 1) siz[(x+c-1)/c]++;
32 }
33 inline void Era(int x) {
34     x = a[x];
35     cnt[x]--;
36     if (!cnt[x]) siz[(x+c-1)/c]--;
37 }
38 inline int Query(int K) {
39     for (int b = 1; ; b++)
40         if (siz[b] < K) K -= siz[b];
41         else {
42             for (int i = (b-1)*c+1; i <= b*c; i++) {
43                 if (cnt[i] > 0) K--;
44                 if (!K) return Hash[i];
45             }
46         }
47 }
48
49 int main(void) {
50     #ifndef ONLINE_JUDGE
51         freopen("xsy1640.in", "r", stdin);
52     #endif
53
54     n = rd(), m = rd();
55     F(i, 1, n) a[i] = rd();
56
57     F(i, 1, n) Hash[++tot] = a[i];
58     sort(Hash+1, Hash+tot+1);
59     tot = unique(Hash+1, Hash+tot+1)-Hash-1;
60     F(i, 1, n) a[i] = lower_bound(Hash+1, Hash+tot+1, a[i]) - Hash;
61
62     c = (int)sqrt(n);
63     F(i, 1, m) {
64         int l = rd(), r = rd(), K = rd();
65         qs[i] = (Qry){l, r, K, i};
66     }
67     sort(qs+1, qs+m+1);
68
69     int mL = 1, mR = 0;
70     F(i, 1, m) {
71         while (mR < qs[i].r)
72             Add(++mR);
73         while (mL > qs[i].l)
74             Add(--mL);
75         while (mR > qs[i].r)
76             Era(mR--);
77         while (mL < qs[i].l)
78             Era(mL++);
79         ans[qs[i].id] = Query(qs[i].K);
80     }
81     F(i, 1, m) printf("%d\n", ans[i]);
82
83     return 0;
84 }