【数据结构 | C++】逆序对

逆序对

问题描述：对于给定的正整数序列a，如果 a[i]>a[j] 且 i < j 则 a[i]与a[j] 为一个逆序对。请问，对于给定的序列，有多少个逆序对？

数据输入：第一行一个正整数n(1<=n<=500000)，表示序列有n个数，第二行n个正整数，每个数不超过n。

数据输出：序列的逆序对的数量

输入示例
sample1:
6
5 4 2 6 3 1
sample2:
3
2 2 1

输出示例
sample1:
11
sample2:
2

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

long long merge_and_count(vector<int>& arr, int left, int mid, int right) {
    int i = left, j = mid + 1;
    vector<int> temp;
    long long count = 0;

    while (i <= mid && j <= right) {
        if (arr[i] <= arr[j]) {
            temp.push_back(arr[i++]);
        } else {
            temp.push_back(arr[j++]);
            count += (mid - i + 1);  // 左边剩余的元素数
        }
    }

    while (i <= mid) temp.push_back(arr[i++]);
    while (j <= right) temp.push_back(arr[j++]);

    for (int k = left; k <= right; ++k) {
        arr[k] = temp[k - left];
    }

    return count;
}

long long merge_sort_and_count(vector<int>& arr, int left, int right) {
    if (left >= right) return 0;

    int mid = left + (right - left) / 2;
    long long count = 0;

    count += merge_sort_and_count(arr, left, mid);
    count += merge_sort_and_count(arr, mid + 1, right);
    count += merge_and_count(arr, left, mid, right);

    return count;
}

int main() {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
    }

    cout << merge_sort_and_count(arr, 0, n - 1) << endl;
    return 0;
}