E. Special Segments of Permutation 题解(树上启发式合并)

[题目链接](E. Special Segments of Permutation)

题目思路

其实就是树上启发式合并的思维

单调栈预处理以\(a[i]\)为最大值的区间，然后每次只计算小区间

玄学一点分析，就是树上启发式合并

代码

#include<bits/stdc++.h>
#define fi first
#define se second
#define debug cout<<"I AM HERE"<<endl;
using namespace std;
typedef long long ll;
const int maxn=2e5+5,inf=0x3f3f3f3f,mod=1e9+7;
const double eps=1e-6;
int n;
int a[maxn],l[maxn],r[maxn];
int pos[maxn];
stack<int> sta;
signed main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        pos[a[i]]=i;
    }
    for(int i=n;i>=1;i--){
        while(!sta.empty()&&a[sta.top()]<=a[i]){
            sta.pop();
        }
        r[i]=sta.empty()?n:sta.top()-1;
        sta.push(i);
    }
    while(!sta.empty()) sta.pop();
    for(int i=1;i<=n;i++){
        while(!sta.empty()&&a[sta.top()]<=a[i]){
            sta.pop();
        }
        l[i]=sta.empty()?1:sta.top()+1;
        sta.push(i);
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        if(i-l[i]<r[i]-i){
            for(int j=l[i];j<i;j++){
                int other=pos[a[i]-a[j]];
                if(other>i&&other<=r[i]){
                    ans++;
                }
            }
        }else{
             for(int j=i+1;j<=r[i];j++){
                int other=pos[a[i]-a[j]];
                if(other<i&&other>=l[i]){
                    ans++;
                }
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}