Lost Cows POJ - 2182 线段树维护

N (2 <= N <= 8,000) cows have unique brands in the range 1…N. In a spectacular display of poor judgment, they visited the neighborhood ‘watering hole’ and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands.

Regrettably, FJ does not have a way to sort them. Furthermore, he’s not very good at observing problems. Instead of writing down each cow’s brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow.

Given this data, tell FJ the exact ordering of the cows.
Input

• Line 1: A single integer, N

• Lines 2…N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on.
  Output

• Lines 1…N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on.
  Sample Input
  5
  1
  2
  1
  0
  Sample Output
  2
  4
  5
  3
  1
  题意：每头牛的身高唯一，现在给定从2~n奶牛前面有多少个奶牛的身高比它矮，还原原来的序列；

倒序遍历一遍，扫描到a，那么肯定是剩余序列的第 a+1 大的数，然后删去这个数，重复上述步骤即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>


ll mul(ll a, ll b, ll mod) {
	ll rt = 0;
	while (b) {
		if (b & 1)rt = (rt + a) % mod;
		b = b / 2;
		a = (a << 1) % mod;
	}
	return rt;
}


ll quickpow(ll a, ll b,ll mod) {
	ll ans = 1, tmp = a;
	while (b > 0) {
		if (b % 2)ans = mul(ans, tmp, mod);
		b = b / 2;
		tmp = mul(tmp, tmp, mod);
	}
	return ans;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}

int smaller[maxn], res[maxn];
struct Tree {
	int l, r, len;
}node[maxn];

void build(int l, int r, int rot) {
	node[rot].l = l;
	node[rot].r = r;
	node[rot].len = r - l + 1;
	if (l == r)return;
	int mid = (l + r) >> 1;
	build(l, mid, rot << 1);
	build(mid + 1, r, rot << 1 | 1);
}

int query(int rot, int k) {
	node[rot].len--;
	if (node[rot].l == node[rot].r) {
		return node[rot].l;
	}
	else if (node[rot<<1].len >= k)return query(rot << 1, k);
	else return query(rot << 1 | 1, k - node[rot<<1].len);
}

int main() {
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	for (int i = 2; i <= n; i++)cin >> smaller[i];
	smaller[1] = 0;
	build(1, n, 1);
	for (int i = n; i >= 1; i--) {
		res[i] = query(1, smaller[i] + 1);
	}
	for (int i = 1; i <= n; i++)
		cout << res[i] << endl;
}