[开源社区openEuler实践】873 div2 D2. Range Sorting (Hard Version)

题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5;
const int mod = 1e9 + 7;
// const int mod = 998244353;
const int N = 1e5;
// int a[505][5005];
// bool vis[505][505];
// char s[505][505];
int a[maxn], b[maxn];
int vis[maxn];
string s;
int n, m;

struct Node{
    int val, id;
    bool operator<(const Node &u)const{
        return val < u.val;
    }
}c[maxn];

int ans[maxn], pre[maxn];

void solve(){
    int res = 0;
    // int q, k;
    // int x;
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
        c[i] = {a[i], i};
    }
    sort(c + 1, c + n + 1);
    set<int> low = {0, n + 1}, high = {0, n + 1};
    for(int i = 1; i <= n; i++){
        low.insert(i);
    }
    for(int len = 1; len <= n; len++){
        res += (len - 1) * (n - len + 1);
    }
    for(int i = n; i >= 1; i--){//从大到小遍历
        int id = c[i].id;
        int val = c[i].val;
        low.erase(id);
        high.insert(id);
        int y = *low.upper_bound(id);//在小于val的数的下标中找在id右边最近的
        int k = *prev(low.upper_bound(id));//小于val的数的下标中左边最近的
        int x = *prev(high.upper_bound(k));//在大于val的数的下标中找在k左边最近的
        res -= max(0LL, (y - id) * (k - x));
    }
    cout << res << '\n';
}
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);

    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}