Arrays(问题转化)

题意：给一个长度为n的序列，将一个非空子区间排序可以得到新序列，问这个序列能得到多少不同序列？

求出有多少子区间，左端值不是最小值，右端值不是最大值
容易发现答案=满足条件子区间个数+1

先用倍增/单调队列 求出对于每个左端点i，右边第一个<ai<script type="math/tex" id="MathJax-Element-1"> lpi，同理处理右端点 rpi
然后用树状数组离线统计

#include<bits/stdc++.h> 
using namespace std;
#define
#define
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#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
#define
#define
#define
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
#define
int a[MAXN],a2[MAXN];
int logg2[MAXN];
class RMQ
{
public:
    int n,a[MAXN];
    void mem(int* _a,int _n=0){n=_n;memcpy(a,_a,sizeof(int)*_n);    }
    int d[MAXN][MAXLog],s[MAXN][MAXLog];
    void RMQ_init()
    {
        Rep(i,n) d[i][0]=s[i][0]=a[i];
        for(int j=1;(1<<j)<=n;j++)
            for(int i=0;i + (1<<j) -1 < n; i++)
            {
                d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);
                s[i][j]=max(s[i][j-1],s[i+(1<<(j-1))][j-1]);
            }
    } 
    int query(int L,int R)
    {
        int k=logg2[R-L+1];
        return min(d[L][k],d[R-(1<<k)+1][k]);
    }
    int query2(int L,int R)
    {
        int k=logg2[R-L+1];
        return max(s[L][k],s[R-(1<<k)+1][k]);
    }
}S;
int lp[MAXN],rp[MAXN],n;
struct BIT{
    ll f[MAXN]; 
    void add(int x,ll v) {
        for(int i=x;i<=n;i+=i&(-i)) 
                f[i]+=v;
    }
    ll qur(int x) {
        ll v=0;
        for(int i=x;i;i-=i&(-i)) 
                v+=f[i];
        return v;
    }
}T;
pi p[MAXN];
int main()
{
//  freopen("A_data.in","r",stdin);
//  freopen(".out","w",stdout);
    n=read();
    Rep(i,n) a[i] = read();
    int cnt=0;
    for(int i=0,j=1;cnt<=n;i++,j*=2) {
        for(int p=1;p<=j&&cnt<=n;++p) logg2[++cnt]=i;
    }

    S.mem(a,n);
    S.RMQ_init();

    Rep(i,n){
        int l=i,r=n-1,ans=n;
        while(l<=r) {
            int m=l+r>>1;
            if (S.query(i,m)<a[i]) ans=m,r=m-1;else l=m+1;
        }
        lp[i]=ans;
    }

    Rep(i,n){
        int l=0,r=i-1,ans=-1;
        while(l<=r) {
            int m=l+r>>1;
            if (S.query2(m,i)>a[i]) ans=m,l=m+1;else r=m-1;
        }
        rp[i]=ans;
    }

    Rep(i,n)  p[i]=mp(rp[i],i);
    sort(p,p+n);

    ll ans=0;
    int nowl=-1;
    MEM(T.f)
    Rep(i,n) {
        int id=p[i].se,nowi=p[i].fi;
        while(nowl<nowi) {
            ++nowl;
            if(lp[nowl]==n) continue;
            T.add(lp[nowl]+1,1);
        }       
        int c=T.qur(id+1) ;
        ans +=c;
    }
    cout<<ans+1<<endl;

    return 0;
}