洛谷 - P4755 Beautiful Pair

传送门：洛谷 - P4755

思路分析

枚举$i$作为区间最大值的贡献，只需要枚举区间的一边，然后查询另一边有几个数满足
以权值较大为根来建立笛卡尔树，可以很好的维护$i$作为区间最大值时的左端点和有端点，查询利用主席树完成

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 1e5+10;

#define int long long

struct node{
    int l,r,sum;
}tree[N*40];

int rt[N],cnt;

#define ls(x) tree[x].l
#define rs(x) tree[x].r
#define mid (l+r>>1)

void pp(int x){
    tree[x].sum = tree[ls(x)].sum + tree[rs(x)].sum;
}

void ins(int l,int r,int &x,int pre,int val){
    x=++cnt;
    tree[x] = tree[pre];
    if(l==r){
        tree[x].sum++;
        return ;
    }
    if(val<=mid) ins(l,mid,ls(x),ls(pre),val);
    else ins(mid+1,r,rs(x),rs(pre),val);
    pp(x);
}

int query(int l,int r,int L,int R,int ql,int qr){
    if(ql<=l && r<=qr){
        return tree[R].sum - tree[L].sum;
    }
    int ans = 0;
    if(ql<=mid) ans += query(l,mid,ls(L),ls(R),ql,qr);
    if(qr>mid) ans += query(mid+1,r,rs(L),rs(R),ql,qr);
    return ans;
}

namespace ditree{

    struct node{
        int ls,rs,val;
    }tree[N];

    int st[N],top=0,root,ans;

    void build(int *a,int n){
        for(int i=1;i<=n;i++){
            int j=0;
            while(top && a[st[top]]<a[i]) j = st[top],top--;
            if(!top) root = i;
            else tree[st[top]].rs = i;
            tree[i].ls = j;
            tree[i].val = a[i];
            st[++top] = i;
        }
    }

    void dfs(int x,int l,int r){
        if(x-l<r-x) for(int i=l;i<=x;i++) ans += query(1,1e9,rt[x-1],rt[r],1,tree[x].val/tree[i].val);
        else for(int i=x;i<=r;i++) ans += query(1,1e9,rt[l-1],rt[x],1,tree[x].val/tree[i].val);
        if(tree[x].ls) dfs(tree[x].ls,l,x-1);
        if(tree[x].rs) dfs(tree[x].rs,x+1,r);
    }
};

int a[N];


signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n;
    read(n);
    for(int i=1;i<=n;i++) {
        read(a[i]);
        ins(1,1e9,rt[i],rt[i-1],a[i]);
    }
    ditree::build(a,n);
    ditree::dfs(ditree::root,1,n);
    printf("%lld\n",ditree::ans);


    return 0;
}