[BZOJ4989] [Usaco2017 Feb]Why Did the Cow Cross the Road（树状数组）

传送门

发现就是逆序对

可以树状数组求出

对于旋转操作，把一个序列最后面一个数移到开头，假设另一个序列的这个数在位置x，那么对答案的贡献 - (n - x) + (x - 1)

#include <cstdio>
#include <cstring>
#include <iostream>
#define N 200011
#define LL long long

using namespace std;

int n;
int a[N], b[N], pos1[N], pos2[N];
LL c[N], ans = 1ll * N * N;

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void add(int x)
{
	for(; x <= n; x += x & -x) c[x]++;
}

inline LL query(int x)
{
	LL ret = 0;
	for(; x; x -= x & -x) ret += c[x];
	return ret;
}

inline void solve()
{
	int i, x;
	LL sum = 0;
	memset(c, 0, sizeof(c));
	for(i = 1; i <= n; i++) pos1[a[i]] = i;
	for(i = 1; i <= n; i++) pos2[b[i]] = i;
	for(i = 1; i <= n; i++)
	{
		x = pos1[b[i]];
		sum += (LL)i - 1 - query(x);
		add(x);
	}
	ans = min(ans, sum);
	for(i = n; i > 1; i--)
	{
		x = pos2[a[i]];
		sum -= n - x;
		sum += x - 1;
		ans = min(ans, sum);
	}
}

int main()
{
	int i;
	n = read();
	for(i = 1; i <= n; i++) a[i] = read();
	for(i = 1; i <= n; i++) b[i] = read();
	solve();
	swap(a, b);
	solve();
	printf("%lld\n", ans);
	return 0;
}