Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3605 [USACO17JAN]Promotion Counting P

P3605 [USACO17JAN]Promotion Counting P

一般这类问题如果是有减意义的(加)可以直接扫描线,否则(最大值)可以考虑线段树合并或者树上启发式合并。

查看代码
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
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
const int N = 1e5 + 10;
int n, m, w[N], tr[N], ans[N];
vector <int> g[N];
void modify (int x, int k)
{
for (; x; x -= x & -x) tr[x] += k;
}
int query (int x)
{
int res = 0;
for (; x <= m; x += x & -x) res += tr[x];
return res;
}
void dfs (int x)
{
ans[x] -= query(w[x] + 1);
for (int i : g[x]) dfs(i);
ans[x] += query(w[x] + 1);
modify(w[x], 1);
}
int main ()
{
read(n);
vector <int> ws;
for (int i = 1; i <= n; ++i)
read(w[i]), ws.push_back(w[i]);
sort(ws.begin(), ws.end());
m = ws.erase(unique(ws.begin(), ws.end()), ws.end()) - ws.begin();
for (int i = 1; i <= n; ++i)
w[i] = upper_bound(ws.begin(), ws.end(), w[i]) - ws.begin();
for (int i = 2, a; i <= n; ++i)
read(a), g[a].push_back(i);
dfs(1);
for (int i = 1; i <= n; ++i)
write(ans[i]), puts("");
return 0;
}