CF1454F Array Partition

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：

#include<bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 200005;
int a[N], st[N][20];
int log2(int x)
{
    int res = -1;
    while (x) { x >>= 1; res++; }
    return res;
}
void init(int n)
{
    for (int i = 0; i < n; i++) st[i][0] = a[i];
    for (int j = 1; (1 << j) < n; j++)
    {
        for (int i = 0; i + (1 << j) - 1 < n; i++)
        {
            st[i][j] = min(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
        }
    }
}
int query(int l, int r)
{
    if (l > r) return INF;
    int p = log2(r - l + 1);
    return min(st[l][p], st[r - (1 << p) + 1][p]);
}
int main()
{
    int t; cin >> t;
    while (t--)
    {
        int n; cin >> n;
        for (int i = 0; i < n; i++) cin >> a[i];
        init(n);
        vector<int> prefix_max(n, 0), suffix_max(n, 0);
        prefix_max[0] = a[0];
        for (int i = 1; i < n; i++) prefix_max[i] = max(prefix_max[i - 1], a[i]);
        suffix_max[n - 1] = a[n - 1];
        for (int i = n - 2; i >= 0; i--) suffix_max[i] = max(suffix_max[i + 1], a[i]);
        int L = -1, R = -1;
        for (int i = n - 1; i >= 0; i--)
        {
            int x = suffix_max[i];
            int l = lower_bound(prefix_max.begin(), prefix_max.begin() + i, x) - prefix_max.begin() + 1;
            int r = upper_bound(prefix_max.begin(), prefix_max.begin() + i, x) - prefix_max.begin();
            if (l > r) continue;
            while (l <= r)
            {
                int m = l + r >> 1;
                int tmp = query(m, i - 1);
                if (tmp < x) l = m + 1;
                else
                {
                    if (tmp == x) { L = m; break; }
                    r = m - 1;
                }
            }
            if (L != -1) { R = i; break; }
        }
        if (L != -1)
        {
            int end = n - R, mid = n - L - end;
            cout << "YES" << endl;
            cout << L << " " << mid << " " << end << endl;
        }
        else cout << "NO" << endl;
    }
    return 0;
}