【JZOJ6342】Tiny Counting

description

---

analysis

• 首先不管(a,b,c,d)重复的情况方案数是正逆序对之积

• 如果考虑(a,b,c,d)有重复，只有四种情况，下面括号括起来表示该位置重复

• 比如({a,(b,c),d})，其中(b=c,S_a<S_b,S_c>S_d)

• 还有({(a,c),b,d})，({a,c,(b,d)})，({c,(a,d),b})，注意前两种不在括号内的数可以互换

• 那么扫两次，用数据结构维护在某个位置前面或后面比该数大或小的数

• 于是拿总方案减掉四种不合法的方案数，这个直接用维护出来的数求

• 由于我很蠢笨树状数组都不会只能用线段树苟且偷生

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 100005
#define ll long long
#define reg register ll
#define max(x,y) ((x>y)?(x):(y))
#define min(x,y) ((x<y)?(x):(y))
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll tr[MAXN<<2],b[MAXN],c[MAXN];
ll premx[MAXN],premn[MAXN],sufmx[MAXN],sufmn[MAXN];
ll n,m,ans,ans1;

struct node
{
	ll x,id;
}a[MAXN];

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline void swap(ll &x,ll &y){ll z=x;x=y,y=z;}
inline bool cmp(node a,node b){return a.x<b.x;}
inline bool cmp1(node a,node b){return a.id<b.id;}
inline void modify(ll t,ll l,ll r,ll x)
{
	if (l==r){++tr[t];return;}
	ll mid=(l+r)>>1;if (x<=mid)modify(t<<1,l,mid,x);
	else modify((t<<1)+1,mid+1,r,x);tr[t]=tr[t<<1]+tr[(t<<1)+1];
}
inline ll query(ll t,ll l,ll r,ll x,ll y)
{
	if (x>y)return 0;ll mid=(l+r)>>1;
	if (l==x && y==r)return tr[t];
	if (y<=mid)return query(t<<1,l,mid,x,y);
	else if (x>mid)return query((t<<1)+1,mid+1,r,x,y);
	else return query(t<<1,l,mid,x,mid)+query((t<<1)+1,mid+1,r,mid+1,y);
}
inline void msort(ll l,ll r)
{
	if (l==r)return;
	ll mid=(l+r)>>1,i=l,j=mid+1,k=l;
	msort(l,mid),msort(mid+1,r);
	while (i<=mid && j<=r)
	{
		if (b[i]<=b[j])c[k++]=b[i++];
		else c[k++]=b[j++],ans+=mid-i+1;
	}
	while (i<=mid)c[k++]=b[i++];
	while (j<=r)c[k++]=b[j++];
	fo(i,l,r)b[i]=c[i];
}
int main()
{
	//freopen("T1.in","r",stdin);
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	n=read();fo(i,1,n)a[i].x=read(),a[i].id=i;
	sort(a+1,a+n+1,cmp);ll i=1,tmp=0;
	while (i<=n){tmp=a[i].x,a[i].x=++m;while (i<n && a[i+1].x==tmp)a[++i].x=m;++i;}
	sort(a+1,a+n+1,cmp1);fo(i,1,n)b[i]=a[i].x;msort(1,n),ans1=ans,ans=0;
	fo(i,1,n)b[i]=a[n-i+1].x;msort(1,n),ans=(ll)ans*ans1;
	fo(i,1,n)premx[i]=query(1,1,n,a[i].x+1,n),premn[i]=query(1,1,n,1,a[i].x-1),modify(1,1,n,a[i].x);
	memset(tr,0,sizeof(tr));
	fd(i,n,1)sufmx[i]=query(1,1,n,a[i].x+1,n),sufmn[i]=query(1,1,n,1,a[i].x-1),modify(1,1,n,a[i].x);
	fo(i,1,n)ans-=premx[i]*sufmx[i]+premn[i]*sufmn[i]+premx[i]*premn[i]+sufmx[i]*sufmn[i];
	printf("%lld
",ans);
	return 0;
}