World is Exploding

题意：

给出一个长为n的序列A，问有多少四元组(a, b, c, d)满足$a \ne b \ne c \ne d, 1 \leq a < b \leq n, 1 \leq c < d \leq n, A_a < A_b, A_c > A_d$.

 

解法：

容斥，考虑用正序对乘以逆序对后，得到的方案中四种不合法情况：

1.$c < d = a < b$

2.$a < b = c < d$

3.$a = c < d$ 且 $a = c < d$

4.$a < b = d$ 且 $c < b = d$

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

#define N 50010
#define LL long long
#define lb(x) (x&(-x))

using namespace std;

int n, tot;
int a0[N], a[N], pre[N], suc[N];

int ask(int sum[], int x)
{
    int ans = 0;
    for(int i = x; i > 0; i -= lb(i))
        ans += sum[i];
    return ans;
}

void add(int sum[], int x, int v)
{
    for(int i = x; i <= tot && i>0;i += lb(i))
        sum[i] += v;
}

LL qsum(int sum[], int l, int r)
{
    if(l > r) return 0;
    return ask(sum, r) - ask(sum, l - 1);
}

int main()
{
    while(~scanf("%d", &n))
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            a0[i] = a[i];
            suc[i] = 0;
            pre[i] = 0;
        }
        sort(a0 + 1, a0 + n + 1);
        tot = 1;
        for(int i = 2; i <= n; i++)
            if(a0[i] != a0[i-1]) a0[++tot] = a0[i];
        for(int i = 1; i <= n; i++)
        {
            a[i] = lower_bound(a0 + 1, a0 + tot + 1, a[i]) - a0;
            add(suc, a[i], 1);
        }
        LL ans = 0, cnt1 = 0, cnt2 = 0;
        for(int i = 1; i <= n; i++)
        {
            add(pre, a[i], 1);
            add(suc, a[i], -1);
            LL pre_higher = qsum(pre, a[i] + 1, tot);
            LL pre_lower = qsum(pre, 1, a[i] - 1);
            LL suc_higher = qsum(suc, a[i] + 1, tot);
            LL suc_lower = qsum(suc, 1, a[i] - 1);
            cnt1 += suc_higher;
            cnt2 += suc_lower;
            ans -= pre_higher * suc_higher;
            ans -= pre_lower * suc_lower;
            ans -= suc_lower * suc_higher;
            ans -= pre_lower * pre_higher;
        }
        ans += cnt1 * cnt2;
        cout << ans << endl;
    }
    return 0;
}