[AcWing 788] 逆序对的数量

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点击查看代码

#include<iostream>

using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int a[N], tmp[N];
ll merge_sort(int l, int r)
{
    if (l >= r) return 0;
    int mid = l + r >> 1;
    ll res = merge_sort(l, mid) + merge_sort(mid + 1, r);
    int i = l, j = mid + 1, k = 0;
    while (i <= mid && j <= r) {
        if (a[i] <= a[j])   tmp[k++] = a[i++];
        else {
            tmp[k++] = a[j++];
            res += mid - i + 1;
        }
    }
    while (i <= mid)    tmp[k++] = a[i++];
    while (j <= r)      tmp[k++] = a[j++];
    for (int i = l, j = 0; i <= r; i++, j++)    a[i] = tmp[j];
    return res;
}
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    printf("%lld", merge_sort(0, n - 1));
    return 0;
}

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1. l >= r 时，直接返回 0；
2. 递归地处理左半边和右半边，并把逆序对的数目加到 res；
3. 归并排序模板，区别之处在于 a[i] > a[j] 时，res += mid - i + 1，代表的是 j 在左半边的逆序对个数；
4. 注意要开 long long；