BZOJ3809 Gty的二逼妹子序列

 1 /**************************************************************
 2     Problem: 3809
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:26248 ms
 7     Memory:25428 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cmath>
12 #include <algorithm>
13  
14 using namespace std;
15 const int N = 100005;
16 const int M = 1000005;
17 const int sqrt_N = 500;
18  
19 int n, Q, sz;
20 int a[N];
21 int B[sqrt_N], w[N], cnt[N];
22 int ans[M];
23  
24 struct data {
25     int l, r, a, b, W;
26      
27     inline bool operator < (const data &x) const {
28         return w[l] == w[x.l] ? r < x.r : w[l] < w[x.l];
29     }
30 } q[M];
31  
32 inline int read() {
33     int x = 0;
34     char ch = getchar();
35     while (ch < ‘0‘ || ‘9‘ < ch)
36         ch = getchar();
37     while (‘0‘ <= ch && ch <= ‘9‘) {
38         x = x * 10 + ch - ‘0‘;
39         ch = getchar();
40     }
41     return x;
42 }
43  
44 int query(int x, int y) {
45     int res = 0, i;
46     if (w[x] == w[y]) {
47         for (i = x; i <= y; ++i)
48             if (cnt[i]) ++res;
49     } else {
50         for (i = w[x] + 1; i < w[y]; ++i)
51             res += B[i];
52         for (i = w[x] * sz; i >= x; --i)
53             if (cnt[i]) ++res;
54         for (i = (w[y] - 1) * sz + 1; i <= y; ++i)
55             if (cnt[i]) ++res;
56     }
57     return res;
58 }
59  
60 inline void add(int c, int del) {
61     if (!cnt[c]) ++B[w[c]];
62     cnt[c] += del;
63     if (!cnt[c]) --B[w[c]];
64 }
65  
66 int main() {
67     int i, l, r;
68     n = read(), Q = read(), sz = (int) sqrt(n / 2);
69     for (i = 1; i <= n; ++i)
70         a[i] = read(), w[i] = (int) (i - 1) / sz + 1;
71     for (i = 1; i <= Q; ++i) {
72         q[i].l = read(), q[i].r = read(), q[i].a = read(), q[i].b = read();
73         q[i].W = i;
74     }
75     sort(q + 1, q + Q + 1);
76      
77     for (i = l = 1, r = 0; i <= Q; ++i) {
78         for (; l < q[i].l; ) add(a[l++], -1);
79         for (; l > q[i].l; ) add(a[--l], 1);
80         for (; r < q[i].r; ) add(a[++r], 1);
81         for (; r > q[i].r; ) add(a[r--], -1);
82         ans[q[i].W] = query(q[i].a, q[i].b);
83     }
84     for (i = 1; i <= Q; ++i)
85         printf("%d\n", ans[i]);
86     return 0;
87 }