NC51101 Lost Cows

题目链接

题目

题目描述

\(N (2 \leq N \leq 8,000)\) 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.

输入描述

• 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.

输出描述

• 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.

示例1

输入

5
1
2
1
0

输出

2
4
5
3
1

题解

知识点：树状数组，倍增，枚举。

首先需要一个事实，对于一个排列，要确定其中某个位置的数具体是多少，可以通过整个排列比它小的数字有多少个（或者比它大的数字有多少个）确定。现在这道题确定了各个位置的数的左边比它小的数的个数，我们只需要知道在它右边有多少个数比它小就行，因此我们从右往左枚举，依次确定数字。

首先用树状数组维护右侧出现的数字，随后需要二分一个 \(x\) 通过比较 \(x - cnt_{<x} -1 < a_i\) 得知 \(x\) 是否小了还是大了，从而找到第一个 \(x - cnt_{<x} -1 = a_i\) 的点。注意，条件不能为 \(x - cnt_{<x} -1 \leq a_i\) ， 因为可能会出现连续一段刚好等于 \(a_i\) 的点，而我们只需要第一个的下标即可，如果用这个条件，我们得到的是最后一个下标。

当然，这个二分条件其实还可以更简单，我们反过来记录右侧没出现的数字，那么 \(cnt_{<x}\) 直接代表左侧比 \(x\) 小的数字个数，那么条件为 \(cnt_{<x} < a_i\) 即可。

另外，二分套树状数组查询的复杂度是对数平方的，并不是最优的。我们可以直接使用树状数组本身的倍增性质进行二分，是最优的对数复杂度。我封装了两个常用的函数，大于等于 lower_bound 和大于 upper_bound 。

这道题查询 \(cnt_{<x} = a_i\) 的第一个点，我们 lower_bound 查询即可。

时间复杂度 \(O(n \log n)\)

空间复杂度 \(O(n)\)

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

template <class T>
class Fenwick {
    int n;
    vector<T> node;

public:
    Fenwick(int _n = 0) { init(_n); }

    void init(int _n) {
        n = _n;
        node.assign(n + 1, T());
    }

    void update(int x, T val) { for (int i = x;i <= n;i += i & -i) node[i] += val; }

    T query(int x) {
        T ans = T();
        for (int i = x;i >= 1;i -= i & -i) ans += node[i];
        return ans;
    }
    T query(int l, int r) { return query(r) - query(l - 1); }

    int lower_bound(T val) {
        int pos = 0;
        for (int i = 1 << __lg(n); i; i >>= 1) {
            if (pos + i <= n && node[pos + i] < val) {
                pos += i;
                val -= node[pos];
            }
        }
        return pos + 1;
    }
    int upper_bound(T val) {
        int pos = 0;
        for (int i = 1 << __lg(n); i; i >>= 1) {
            if (pos + i <= n && node[pos + i] <= val) {
                pos += i;
                val -= node[pos];
            }
        }
        return pos + 1;
    }
};

int a[8007];
int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    for (int i = 2;i <= n;i++) cin >> a[i];
    Fenwick<int> fw(n);
    for (int i = 1;i <= n;i++) fw.update(i, 1);

    vector<int> ans(n + 1);
    for (int i = n;i >= 1;i--) {
        ans[i] = fw.lower_bound(a[i] + 1);
        fw.update(ans[i], -1);
    }
    for (int i = 1;i <= n;i++) cout << ans[i] << '\n';
    return 0;
}