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 = ;
for(int i = x; i > ; i -= lb(i))
ans += sum[i];
return ans;
} void add(int sum[], int x, int v)
{
for(int i = x; i <= tot && i>;i += lb(i))
sum[i] += v;
} LL qsum(int sum[], int l, int r)
{
if(l > r) return ;
return ask(sum, r) - ask(sum, l - );
} int main()
{
while(~scanf("%d", &n))
{
for(int i = ; i <= n; i++)
{
scanf("%d", &a[i]);
a0[i] = a[i];
suc[i] = ;
pre[i] = ;
}
sort(a0 + , a0 + n + );
tot = ;
for(int i = ; i <= n; i++)
if(a0[i] != a0[i-]) a0[++tot] = a0[i];
for(int i = ; i <= n; i++)
{
a[i] = lower_bound(a0 + , a0 + tot + , a[i]) - a0;
add(suc, a[i], );
}
LL ans = , cnt1 = , cnt2 = ;
for(int i = ; i <= n; i++)
{
add(pre, a[i], );
add(suc, a[i], -);
LL pre_higher = qsum(pre, a[i] + , tot);
LL pre_lower = qsum(pre, , a[i] - );
LL suc_higher = qsum(suc, a[i] + , tot);
LL suc_lower = qsum(suc, , a[i] - );
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 ;
}
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