AcWing244. Mysterious cow (tree array + two points)

AcWing 244. Mysterious Cow ( Two Points )

Ideas

Initialize all numbers to 1, which means they have not been used

Calculate from back to front to find the number of the first k that has not been used. The smallest x that makes sum(x) == k is the answer.

Then set this number to 0 to indicate that it has been used

There are n cows, their heights are known to be 1~n and they are all different, but the specific height of each cow is not known.

Now these n cows stand in a row, it is known that there is $A_{i} in$ front of the i-th cow $A_{i}$ The cow is lower than it. Find the height of each cow.

Input format

Line 1: Enter the integer n.

Line 2...n: Enter an integer $A_{i}$ per line $A_{i}$ , The i-th line means that there is $A_{i} in$ front of the i-th cow $A_{i}$ The head of cattle is lower than it.
(Note: Because there is no cow in front of the first cow, it is not listed)

Output format

The output contains n lines, and each line outputs an integer representing the height of the cow.

The ith line outputs the height of the ith cow.

data range

$1≤n≤10^{5}$

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<string>
#include<cstring>
#include<map>
#include<vector>
#include<set>
#include<stack>
#include<algorithm>
#include<vector>
#include<utility>
#include<deque>
#include<unordered_map>
#define INF 0x3f3f3f3f
#define mod 1000000007
#define endl '\n'
#define eps 1e-6
inline int gcd(int a, int b) {
    
     return b ? gcd(b, a % b) : a; }
inline int lowbit(int x) {
    
     return x & -x; }


using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
const int N = 100010;
int tr[N], h[N], ans[N];

int n;
void add(int x, int c) {
    
    
	for (int i = x; i <= n; i += lowbit(i))tr[i] += c;
}

int query(int x) {
    
    
	int res = 0;
	for (int i = x; i; i -= lowbit(i))res += tr[i];
	return res;
}

int check(int x) {
    
    
	int l = 1, r = n;
	while (l < r) {
    
    
		int mid = l + r >> 1;
		if (query(mid) >= x)r = mid;
		else l = mid + 1;
	}
	add(r, -1);
	return r;
}
int main() {
    
    
	cin >> n;
	for (int i = 2; i <= n; ++i)scanf("%d", &h[i]);

	for (int i = 1; i <= n; ++i)tr[i] = lowbit(i);
	
	for (int i = n; i; --i) {
    
    
		int k = h[i] + 1;
		ans[i] = check(k);
		
	}
	for (int i = 1; i <= n; ++i)printf("%d\n", ans[i]);
	return 0;
}