滑动窗口逆序对(Sliding Window Inversions) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, k;
    if (!(cin >> n >> k)) return 0;
    vector<int> a(n), vals(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        vals[i] = a[i];
    }
    sort(vals.begin(), vals.end());
    vals.erase(unique(vals.begin(), vals.end()), vals.end());
    int m = (int)vals.size();
    for (int i = 0; i < n; ++i) {
        a[i] = (int)(lower_bound(vals.begin(), vals.end(), a[i]) - vals.begin()) + 1;
    }
    vector<int> bit(m + 1, 0);
    auto add = [&](int idx, int v) {
        for (int i = idx; i <= m; i += i & -i) bit[i] += v;
    };
    auto query = [&](int idx) {
        int s = 0;
        for (int i = idx; i > 0; i -= i & -i) s += bit[i];
        return s;
    };
    long long inv = 0;
    int curSz = 0;
    for (int i = 0; i < n; ++i) {
        inv += curSz - query(a[i]);
        add(a[i], 1);
        ++curSz;
        if (curSz > k) {
            int rem = a[i - k];
            add(rem, -1);
            --curSz;
            inv -= query(rem - 1);
        }
        if (i >= k - 1) {
            cout << inv << (i + 1 == n ? "\n" : " ");
        }
    }
    return 0;
}