洛谷p1637三元上升子序列

题目链接

求序列中 i < j < k 且 a [ i ] < a [ j ] < a [ k ] 的 三 元 组 数 量 i< j<k且a[i]<a[j]<a[k]的三元组数量 i<j<k且a[i]<a[j]<a[k]的三元组数量

类似于线段树求逆序对，前后跑两遍即可。

AC代码：

#define MAX 500010
struct node
{
    int k, l, r, sum;
} tr[4 * MAX];
int a[MAX];
int b[MAX];

void pushup(int k)
{
    tr[k].sum = tr[k * 2].sum + tr[k * 2 + 1].sum;
}
void build(int k, int l, int r)
{
    tr[k].l = l;
    tr[k].r = r;
    if (l == r)
    {
        tr[k].sum = 0;
        return;
    }
    int mid = l + r >> 1;
    build(k * 2, l, mid);
    build(k * 2 + 1, mid + 1, r);
    pushup(k);
}

void insert(int k, int w)
{
    if (tr[k].l == tr[k].r)
    {
        tr[k].sum++;
        return;
    }
    int mid = tr[k].l + tr[k].r >> 1;
    if (w <= mid)
        insert(k * 2, w);
    else
        insert(k * 2 + 1, w);
    pushup(k);
}
int query(int k, int l, int r)
{
    if (tr[k].l == l && tr[k].r == r)
        return tr[k].sum;
    int mid = tr[k].l + tr[k].r >> 1;
    if (r <= mid)
        return query(k * 2, l, r);
    else if (l > mid)
        return query(k * 2 + 1, l, r);
    else
        return (query(k * 2, l, mid) + query(k * 2 + 1, mid + 1, r));
}
int bef[MAX], aft[MAX];
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        b[i] = a[i];
    }
    sort(b + 1, b + 1 + n);
    int len = unique(b + 1, b + 1 + n) - b;
    for (int i = 1; i <= n; i++)
        a[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;
    build(1, 1, n);
    for (int i = 1; i <= n; i++)
    {
        insert(1, a[i]);
        if (a[i] > 1)
            bef[i] = query(1, 1, a[i] - 1);
    }

    build(1, 1, n);
    for (int i = n; i >= 1; i--)
    {
        insert(1, a[i]);
        if (a[i] < n)
            aft[i] = query(1, a[i] + 1, n);
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++)
        ans += aft[i] * bef[i];
    printf("%lld\n", ans);

    return 0;
}