Dylans loves sequence HDU - 5273

154 阅读 0 评论 102 点赞

我是靠谱客的博主自信水池，这篇文章主要介绍Dylans loves sequence HDU - 5273，现在分享给大家，希望可以做个参考。

$朴素莫队加树状数组$
$复杂度为 n 根号 n l o g n 显然对于 1 e 5 的数据量超时了$
$以后有空再学二次离线莫队$

复制代码

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int N = 1e5 + 10;
int a[N], c[N], ans[N];
int n, m;
int siz;
struct node{
int l, r, k;
}q[N];
vector<int> vec;
inline int read() {
int s = 0, w = 1;
char ch = getchar();
while (ch<'0' || ch>'9') { if (ch == '-')w = -1; ch = getchar(); }
while (ch >= '0'&&ch <= '9') s = s * 10 + ch - '0', ch = getchar();
return s * w;
}
bool cmp1(node a, node b)
{
if(a.l/siz == b.l/siz) return a.r < b.r;
return a.l/siz < b.l/siz;
}
int lowbit(int x){
return x & -x;
}
void add(int x, int v)
{
for(int i = x; i <= n; i += lowbit(i))
{
c[i] += v;
}
}
int query(int x)
{
int res = 0;
for(int i = x; i > 0; i -= lowbit(i))
{
res += c[i];
}
return res;
}
int find(int x)
{
int l = 0, r = vec.size();
while(l < r)
{
int mid = l + r >> 1;
if(vec[mid] >= x)
{
r = mid;
}
else l = mid + 1;
}
return l + 1;
}
int main(){
n = read();
m = read();
siz = (int)sqrt(n);
for(int i = 1; i <= n; i ++)
{
a[i] = read();
vec.push_back(a[i]);
}
sort(vec.begin(), vec.end());
vec.erase(unique(vec.begin(), vec.end()), vec.end());
int p = vec.size();
for(int i = 0; i < m; i ++)
{
q[i].l = read();
q[i].r = read();
q[i].k = i;
}
sort(q, q+m, cmp1);
int l = 1, r = 0;
int res = 0;
for(int i = 0; i < m; i ++)
{
while(r < q[i].r)
{
r ++;
int t = find(a[r]);
res += (query(p)-query(t));
add(t, 1);
}
while(r > q[i].r)
{
int t = find(a[r]);
if(c[t])
{
res -= (query(p)-query(t));
add(t, -1);
}
r --;
}
while(l < q[i].l)
{
int t = find(a[l]);
if(c[t])
{
res -= query(t-1);
add(t, -1);
}
l ++;
}
while(l > q[i].l)
{
l --;
int t = find(a[l]);
res += query(t-1);
add(t, 1);
}
ans[q[i].k] = res;
}
for(int i = 0; i < m; i ++)
{
printf("%dn", ans[i]);
}
return 0;
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int N = 1e5 + 10;
int a[N], c[N], ans[N];
int n, m;
int siz;
struct node{
int l, r, k;
}q[N];
vector<int> vec;
inline int read() {
int s = 0, w = 1;
char ch = getchar();
while (ch<'0' || ch>'9') { if (ch == '-')w = -1; ch = getchar(); }
while (ch >= '0'&&ch <= '9') s = s * 10 + ch - '0', ch = getchar();
return s * w;
}
bool cmp1(node a, node b)
{
if(a.l/siz == b.l/siz) return a.r < b.r;
return a.l/siz < b.l/siz;
}
int lowbit(int x){
return x & -x;
}
void add(int x, int v)
{
for(int i = x; i <= n; i += lowbit(i))
{
c[i] += v;
}
}
int query(int x)
{
int res = 0;
for(int i = x; i > 0; i -= lowbit(i))
{
res += c[i];
}
return res;
}
int find(int x)
{
int l = 0, r = vec.size();
while(l < r)
{
int mid = l + r >> 1;
if(vec[mid] >= x)
{
r = mid;
}
else l = mid + 1;
}
return l + 1;
}
int main(){
n = read();
m = read();
siz = (int)sqrt(n);
for(int i = 1; i <= n; i ++)
{
a[i] = read();
vec.push_back(a[i]);
}
sort(vec.begin(), vec.end());
vec.erase(unique(vec.begin(), vec.end()), vec.end());
int p = vec.size();
for(int i = 0; i < m; i ++)
{
q[i].l = read();
q[i].r = read();
q[i].k = i;
}
sort(q, q+m, cmp1);
int l = 1, r = 0;
int res = 0;
for(int i = 0; i < m; i ++)
{
while(r < q[i].r)
{
r ++;
int t = find(a[r]);
res += (query(p)-query(t));
add(t, 1);
}
while(r > q[i].r)
{
int t = find(a[r]);
if(c[t])
{
res -= (query(p)-query(t));
add(t, -1);
}
r --;
}
while(l < q[i].l)
{
int t = find(a[l]);
if(c[t])
{
res -= query(t-1);
add(t, -1);
}
l ++;
}
while(l > q[i].l)
{
l --;
int t = find(a[l]);
res += query(t-1);
add(t, 1);
}
ans[q[i].k] = res;
}
for(int i = 0; i < m; i ++)
{
printf("%dn", ans[i]);
}
return 0;
}