OpenJudge 求重要逆序对数

https://blog.csdn.net/mrvector/article/details/81090165

【题解】

         方法与求逆序对的个数类似，用归并排序分治求解。不同之处在于添加了一个虚拟指针pointer。

【代码】

#include <iostream>
using namespace std;
#define maxn 200005

int s[maxn], temp[maxn];
long long ans;

void Merge(int left, int right, int mid)
{
    int i = left, j = mid + 1, k = left;
    int pointer = left;
    while (i <= mid && j <= right) {
        if (s[i] > s[j]) {
            temp[k++] = s[j];
            while (pointer <= mid && s[pointer] <= 2 * s[j])
                pointer++;
            if (pointer < mid + 1)
                ans += mid - pointer + 1;
            j++;
        }
        else {
            temp[k++] = s[i];
            i++;
        }
    }
    while (i <= mid) 
        temp[k++] = s[i++];
    while (j <= right)
        temp[k++] = s[j++];
    for (int i = left; i <= right; i++)
        s[i] = temp[i];
}

void MergeSort(int left, int right)
{
    int mid = (left + right) >> 1;
    if (left < right) {
        MergeSort(left, mid);
        MergeSort(mid + 1, right);
        Merge(left, right, mid);
    }
}

int main()
{
    int N;
    cin >> N;
    for (int i = 0; i < N; i++)
        cin >> s[i];
    MergeSort(0, N - 1);
    cout << ans << endl;
    //system("pause");
    return 0;
}