三元上升子序列（洛谷P1637）

@TOC

三元上升子序列

洛谷P1637

主要思路

树状数组维护一个桶，记录这个数出现的次数

1. 先对数组去重，再进行数据离散化（这个比较坑，一开始没想到）
2. 正向遍历数组，先查找并 $f_i$ 记录比 $a_i$ 小的数出现的次数，再将 $a_i$ 插入树状数组
3. 清空树状数组
4. 反向遍历数组，先查找比 $a_i$ 大的数出现的次数与 $f_i$ 相乘， $ans$ 累加，再将 $a_i$ 插入树状数组

主要代码

signed main()
{
    cin >> n;
    for (register int i = 1; i <= n; ++i)
    {
        cin >> a[i];
        b[i] = a[i];
    }
    sort(b + 1, b + n + 1);
    int op = unique(b + 1, b + n + 1) - b - 1;
    for (int i = 1; i <= n; i++)
    {
        a[i] = lower_bound(b + 1, b + op + 1, a[i]) - b;
    }
    //for(int i = 1; i <= n; i++)cout << a[i] << endl;
    for (register int i = 1; i <= n; i++)
    {

        if (a[i] > 1)
            f[i] = sum(a[i] - 1);
        add(a[i]);
    }
    memset(c, 0, sizeof c);
    for (register int i = n; i >= 1; i--)
    {

        if (a[i] < n)
            ans += (sum(n) - sum(a[i])) * f[i];
        add(a[i]);
    }
    cout << ans << endl;
    return 0;
}

$AC code$

前面的都是快读，代码没那么长，如有需要可拿去用

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int maxn = 1e6 + 5;
int n, m, c[maxn], a[maxn], ans, b[maxn];
int f[maxn];
namespace Fastio
{
    struct Reader
    {
        template <typename T>
        Reader &operator>>(T &x)
        {
            char c = getchar();
            T f = 1;
            x = 0;
            while (c < '0' || c > '9')
            {
                if (c == '-')
                    f = -1;
                c = getchar();
            }
            while (c >= '0' && c <= '9')
                x = x * 10 + (c - '0'), c = getchar();
            x *= f;
            return *this;
        }
        Reader &operator>>(char &c)
        {
            c = getchar();
            while (c == ' ' || c == '\n')
            {
                c = getchar();
            }
            return *this;
        }
        Reader &operator>>(char *str)
        {
            int len = 0;
            char c = getchar();
            while (c == ' ' || c == '\n')
            {
                c = getchar();
            }
            while (c != ' ' && c != '\n' && c != '\r')
            { // \r\n in windows
                str[len++] = c;
                c = getchar();
            }
            str[len] = '\0';
            return *this;
        }
        Reader() {}
    } cin;
    char endl = '\n';
    struct Writer
    {
        template <typename T>
        Writer &operator<<(T x)
        {
            if (x == 0)
            {
                putchar('0');
                return *this;
            }
            if (x < 0)
                putchar('-'), x = -x;
            static int sta[45];
            int top = 0;
            while (x)
                sta[++top] = x % 10, x /= 10;
            while (top)
                putchar(sta[top] + '0'), --top;
            return *this;
        }
        Writer &operator<<(char c)
        {
            putchar(c);
            return *this;
        }
        Writer &operator<<(char *str)
        {
            int cur = 0;
            while (str[cur])
                putchar(str[cur++]);
            return *this;
        }
        Writer() {}
    } cout;
#define cin Fastio::cin
#define cout Fastio::cout
#define endl Fastio::endl
}

inline int lowbit(int x)
{
    return (x & -x);
}

inline void add(register int x)
{
    for (; x <= 1000000; x += lowbit(x))
        c[x]++;
}

inline int sum(register int x)
{
    register int s = 0;
    for (; x; x -= lowbit(x))
        s += c[x];
    return s;
}

signed main()
{
    cin >> n;
    for (register int i = 1; i <= n; ++i)
    {
        cin >> a[i];
        b[i] = a[i];
    }
    sort(b + 1, b + n + 1);
    int op = unique(b + 1, b + n + 1) - b - 1;
    for (int i = 1; i <= n; i++)
    {
        a[i] = lower_bound(b + 1, b + op + 1, a[i]) - b;
    }
    //for(int i = 1; i <= n; i++)cout << a[i] << endl;
    for (register int i = 1; i <= n; i++)
    {

        if (a[i] > 1)
            f[i] = sum(a[i] - 1);
        add(a[i]);
    }
    memset(c, 0, sizeof c);
    for (register int i = n; i >= 1; i--)
    {

        if (a[i] < n)
            ans += (sum(n) - sum(a[i])) * f[i];
        add(a[i]);
    }
    cout << ans << endl;
    return 0;
}