平衡照片(Balanced Photo G) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct BIT {
    int n;
    vector<int> b;
    BIT(int n) : n(n), b(n + 1, 0) {}
    void add(int idx) {
        for (int i = idx; i <= n; i += i & -i) ++b[i];
    }
    int query(int idx) const {
        int s = 0;
        for (int i = idx; i > 0; i -= i & -i) s += b[i];
        return s;
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> h(n), vals(n);
    for (int i = 0; i < n; ++i) {
        cin >> h[i];
        vals[i] = h[i];
    }
    sort(vals.begin(), vals.end());
    vals.erase(unique(vals.begin(), vals.end()), vals.end());
    int m = (int)vals.size();
    for (int i = 0; i < n; ++i) {
        h[i] = (int)(lower_bound(vals.begin(), vals.end(), h[i]) - vals.begin()) + 1;
    }
    vector<int> L(n), R(n);
    BIT bitL(m), bitR(m);
    for (int i = 0; i < n; ++i) {
        L[i] = i - bitL.query(h[i]);
        bitL.add(h[i]);
    }
    for (int i = n - 1; i >= 0; --i) {
        R[i] = (n - 1 - i) - bitR.query(h[i]);
        bitR.add(h[i]);
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
        int mn = min(L[i], R[i]);
        int mx = max(L[i], R[i]);
        if (mx > 2 * mn) ++ans;
    }
    cout << ans << "\n";
    return 0;
}