糖果(Candy) 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 SegTree {
    int n;
    vector<long long> sum;
    vector<int> mx, lz;
    SegTree(int n) : n(n), sum(4 * n + 5, 0), mx(4 * n + 5, 0), lz(4 * n + 5, -1) {}

    void apply(int u, int l, int r, int v) {
        sum[u] = 1LL * (r - l + 1) * v;
        mx[u] = v;
        lz[u] = v;
    }
    void push(int u, int l, int r) {
        if (lz[u] != -1) {
            int mid = (l + r) / 2;
            apply(u * 2, l, mid, lz[u]);
            apply(u * 2 + 1, mid + 1, r, lz[u]);
            lz[u] = -1;
        }
    }
    void pull(int u) {
        sum[u] = sum[u * 2] + sum[u * 2 + 1];
        mx[u] = max(mx[u * 2], mx[u * 2 + 1]);
    }
    void build(int u, int l, int r, const vector<int>& init) {
        if (l == r) {
            sum[u] = mx[u] = init[l];
            return;
        }
        int mid = (l + r) / 2;
        build(u * 2, l, mid, init);
        build(u * 2 + 1, mid + 1, r, init);
        pull(u);
    }
    int find_first_gt(int u, int l, int r, int x) {
        if (mx[u] <= x) return n + 1;
        if (l == r) return l;
        push(u, l, r);
        int mid = (l + r) / 2;
        if (mx[u * 2] > x) return find_first_gt(u * 2, l, mid, x);
        return find_first_gt(u * 2 + 1, mid + 1, r, x);
    }
    void assign(int u, int l, int r, int ql, int qr, int v) {
        if (ql <= l && r <= qr) {
            apply(u, l, r, v);
            return;
        }
        push(u, l, r);
        int mid = (l + r) / 2;
        if (ql <= mid) assign(u * 2, l, mid, ql, qr, v);
        if (qr > mid) assign(u * 2 + 1, mid + 1, r, ql, qr, v);
        pull(u);
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    vector<int> cnt(n + 2, 0), init(n + 1, 0);
    int curMex = 0;
    for (int i = 1; i <= n; ++i) {
        if (a[i] <= n + 1) cnt[a[i]]++;
        while (cnt[curMex] > 0) ++curMex;
        init[i] = curMex;
    }
    vector<int> last(n + 2, n + 1), nxt(n + 1, n + 1);
    for (int i = n; i >= 1; --i) {
        if (a[i] <= n + 1) {
            nxt[i] = last[a[i]];
            last[a[i]] = i;
        }
    }
    SegTree st(n);
    st.build(1, 1, n, init);
    long long ans = 0;
    for (int l = 1; l <= n; ++l) {
        ans += st.sum[1];
        st.assign(1, 1, n, l, l, 0);
        int x = a[l];
        if (x <= n) {
            int pos = max(l + 1, st.find_first_gt(1, 1, n, x));
            int r = nxt[l] - 1;
            if (pos <= r) {
                st.assign(1, 1, n, pos, r, x);
            }
        }
    }
    cout << ans << "\n";
    return 0;
}