3809: Gty的二逼妹子序列

連結

分析：

　　和這道AHOI2013 作業差不多。權值是1~n的，所以對權值進行分塊。$O(1)$修改，$O(\sqrt n)$查詢。

程式碼：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<cmath>
 6 #include<cctype>
 7 #include<set>
 8 #include<queue>
 9
 
 #include<vector>
10 #include<map>
11 using namespace std;
12 typedef long long LL;
13 
14 inline int read() {
15     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
16     for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
17 }
18 
19 const int N = 100005
 
;
20 
21 int bel[N], sum[N], a[N], Ans[N * 10], cnt[N], B;
22 struct Que{ 
23     int l, r, a, b, id;
24     bool operator < (const Que &A) const {
25         return bel[l] == bel[A.l] ? r < A.r : bel[l] < bel[A.l];
26     }
27 }Q[N * 10];
28 
29 inline void add(int x) {
30     cnt[x] ++;
 
31     if (cnt[x] == 1) sum[bel[x]] ++;
32 }
33 inline void del(int x) {
34     cnt[x] --;
35     if (cnt[x] == 0) sum[bel[x]] --;
36 }
37 inline int query(int l,int r) {
38     int res = 0;
39     for (int i = l, lim = min(r, bel[l] * B); i <= lim; ++i) res += (cnt[i] > 0);
40     if (bel[l] != bel[r]) 
41         for (int i = (bel[r] - 1) * B + 1; i <= r; ++i) res += (cnt[i] > 0);
42     for (int i = bel[l] + 1; i <= bel[r] - 1; ++i) res += sum[i];
43     return res;
44 }
45 int main() { 
46     int n = read(), m = read(); B = sqrt(n);
47     for (int i = 1; i <= n; ++i) a[i] = read();
48     for (int i = 1; i <= m; ++i) 
49         Q[i].l = read(), Q[i].r = read(), Q[i].a = read(), Q[i].b = read(), Q[i].id = i;
50     for (int i = 1; i <= n; ++i) bel[i] = (i - 1) / B + 1;
51     sort(Q + 1, Q + m + 1);
52     int L = 1, R = 0;
53     for (int i = 1; i <= m; ++i) {
54         while (L > Q[i].l) add(a[--L]);
55         while (R < Q[i].r) add(a[++R]);
56         while (L < Q[i].l) del(a[L++]);
57         while (R > Q[i].r) del(a[R--]);
58         Ans[Q[i].id] = query(Q[i].a, Q[i].b);
59     }
60     for (int i = 1; i <= m; ++i) printf("%d\n",Ans[i]);
61     return 0;
62 }