思路：

枚举最后一部分，根据最后一部分的最大值确定若干个可能的第一部分，再从这些候选中二分查找确定最终答案。利用了前缀最大值和子段最小值的单调性。

实现：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int INF = 0x3f3f3f3f;
 4 const int N = 200005;
 5 int a[N], st[N][20];
 6 int log2(int x)
 7 {
 8     int res = -1;
 9     while (x) { x >>= 1; res++; }
10     return res;
11 }
12 void init(int n)
13 {
14     for (int i = 0; i < n; i++) st[i][0] = a[i];
15     for (int j = 1; (1 << j) < n; j++)
16     {
17         for (int i = 0; i + (1 << j) - 1 < n; i++)
18         {
19             st[i][j] = min(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
20         }
21     }
22 }
23 int query(int l, int r)
24 {
25     if (l > r) return INF;
26     int p = log2(r - l + 1);
27     return min(st[l][p], st[r - (1 << p) + 1][p]);
28 }
29 int main()
30 {
31     int t; cin >> t;
32     while (t--)
33     {
34         int n; cin >> n;
35         for (int i = 0; i < n; i++) cin >> a[i];
36         init(n);
37         vector<int> prefix_max(n, 0), suffix_max(n, 0);
38         prefix_max[0] = a[0];
39         for (int i = 1; i < n; i++) prefix_max[i] = max(prefix_max[i - 1], a[i]);
40         suffix_max[n - 1] = a[n - 1];
41         for (int i = n - 2; i >= 0; i--) suffix_max[i] = max(suffix_max[i + 1], a[i]);
42         int L = -1, R = -1;
43         for (int i = n - 1; i >= 0; i--)
44         {
45             int x = suffix_max[i];
46             int l = lower_bound(prefix_max.begin(), prefix_max.begin() + i, x) - prefix_max.begin() + 1;
47             int r = upper_bound(prefix_max.begin(), prefix_max.begin() + i, x) - prefix_max.begin();
48             if (l > r) continue;
49             while (l <= r)
50             {
51                 int m = l + r >> 1;
52                 int tmp = query(m, i - 1);
53                 if (tmp < x) l = m + 1;
54                 else
55                 {
56                     if (tmp == x) { L = m; break; }
57                     r = m - 1;
58                 }
59             }
60             if (L != -1) { R = i; break; }
61         }
62         if (L != -1)
63         {
64             int end = n - R, mid = n - L - end;
65             cout << "YES" << endl;
66             cout << L << " " << mid << " " << end << endl;
67         }
68         else cout << "NO" << endl;
69     }
70     return 0;
71 }