Codeforces Round 892 (Div.2)

A. United We Stand

题解

• 赛时想复杂了
• 题目要求我们保证数组\(c\)中的数不是数组\(b\)中任意一个数的因子
• 我们考虑将最小值置于数组\(b\)即可

const int N = 2e5 + 10, M = 4e5 + 10;

int n;
int a[N];

void solve()
{
    cin >> n;
    int mi = INF;
    for (int i = 1; i <= n; ++i)
    {
        cin >> a[i];
        mi = min(a[i], mi);
    }
    vector<int> b, c;
    for (int i = 1; i <= n; ++i)
    {
        if (a[i] == mi)
            b.push_back(a[i]);
        else
            c.push_back(a[i]);
    }
    if (b.size() == 0 || c.size() == 0)
        cout << -1 << endl;
    else
    {
        cout << b.size() << " " << c.size() << endl;
        for (auto x : b)
            cout << x << " ";
        cout << endl;
        for (auto x : c)
            cout << x << " ";
        cout << endl;
    }
}

B. Olya and Game with Arrays

题解

• 我们考虑贪心
• 对于所有数组中的最小值来说，不管我们怎么移动，他对答案的贡献不变
• 所以我们考虑将每个数组中的最小值移动到某一个数组中，使得次小值对答案产生贡献
• 我们如何确定被移动到的是哪一个数组呢
• 只要对次小值排序即可，对于存在最小的次小值的数组就是被移动到的数组

const int N = 2e5 + 10, M = 4e5 + 10;

int n;
vector<int> vec[N];

void solve()
{
    cin >> n;
    int mi = INF;
    vector<int> b;
    for (int i = 1; i <= n; ++i)
    {
        int k;
        cin >> k;
        vec[i].clear();
        for (int j = 1; j <= k; ++j)
        {
            int x;
            cin >> x;
            vec[i].push_back(x);
        }
        sort(all(vec[i]));
        b.push_back(vec[i][1]);
        mi = min(mi, vec[i][0]);
    }
    sort(all(b));
    int ans = 0;
    ans += mi;
    for (int i = 1; i < b.size(); ++i)
        ans += b[i];
    cout << ans << endl;
}

C. Another Permutation Problem

题解

• 打表找规律
• 发现除了\(n=2\)时都是先升序后降序的形式，例如\([1, 2,3,4,8,7,6,5]\)
• \(O(n^2)\)枚举在哪个位置开始降序即可

const int N = 2e5 + 10, M = 4e5 + 10;

int n;
int a[N], b[N], c[N];

void solve()
{
    cin >> n;
    if (n == 2)
    {
        cout << 2 << endl;
        return;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i)
    {
        int mx = 0, sum = 0;
        for (int j = 1; j <= i; ++j)
        {
            sum += j * j;
            mx = max(mx, j * j);
        }
        int t = n;
        for (int j = i + 1; j <= n; ++j)
        {
            sum += t * j;
            mx = max(mx, t * j);
            --t;
        }
        ans = max(ans, sum - mx);
    }
    cout << ans << endl;
}

D. Andrey and Escape from Capygrad

\(1 \leq n, q\leq 2e5\)

题解

• 显然越向右移动越好，所以我们肯定选择\(l_i \leq x \leq b_i\)中较大的\(b_i\)传送，因为只有\(b_j > l_i\)时才能传送，所以实际上的区间为\([l_i, b_i]\)
• 所以问题转化为给定\(n\)个区间\([l_i,b_i]\)，对于数轴上一点\(x > 0\)，对于左端点\(l_i \leq x\)的区间，我们找到其最大的右端点
• 所以我们先考虑将区间合并\(O(nlogn)\)
• 然后离散化区间左端点和询问点
• 然后线段树在左端点处单点修改加上右端点的值即可，维护区间最大值

const int N = 4e5 + 10, M = 4e5 + 10;

int n, q;
struct node
{
    int l, r, a, b;
    bool operator<(const node &t) const
    {
        return l < t.l;
    }
} line[N];
int l[N], r[N], tot, qry[N];

struct info
{
    int mx;
    friend info operator+(const info &a, const info &b)
    {
        info c;
        c.mx = max(a.mx, b.mx);
        return c;
    }
};
struct SEG
{
    info val;
} seg[N << 2];

void up(int id)
{
    seg[id].val = seg[lson].val + seg[rson].val;
}

void build(int id, int l, int r)
{
    if (l == r)
    {
        seg[id].val.mx = 0;
        return;
    }
    int mid = l + r >> 1;
    build(lson, l, mid);
    build(rson, mid + 1, r);
    up(id);
}

void change(int id, int l, int r, int x, int val)
{
    if (l == r)
    {
        seg[id].val.mx = val;
        return;
    }
    int mid = l + r >> 1;
    if (x <= mid)
        change(lson, l, mid, x, val);
    else
        change(rson, mid + 1, r, x, val);
    up(id);
}

info query(int id, int l, int r, int ql, int qr)
{
    if (ql <= l && r <= qr)
    {
        return seg[id].val;
    }
    int mid = l + r >> 1;
    if (qr <= mid)
        return query(lson, l, mid, ql, qr);
    else if (ql > mid)
        return query(rson, mid + 1, r, ql, qr);
    else
        return query(lson, l, mid, ql, qr) + query(rson, mid + 1, r, ql, qr);
}

void solve()
{
    cin >> n;
    for (int i = 1; i <= n; ++i)
    {
        int l, r, a, b;
        cin >> l >> r >> a >> b;
        line[i] = {l, r, a, b};
    }
    // 区间合并
    sort(line + 1, line + n + 1);
    tot = 0;
    int st = line[1].l, ed = line[1].b;
    line[n + 1] = {INF, INF, INF, INF};
    for (int i = 2; i <= n + 1; ++i)
    {
        auto [L, R, a, b] = line[i];
        if (L > ed)
        {
            tot++;
            l[tot] = st, r[tot] = ed;
            st = L, ed = b;
        }
        else
            ed = max(ed, b);
    }
    vector<int> vec;
    auto find = [&](int x)
    {
        return lower_bound(all(vec), x) - vec.begin() + 1;
    };
    for (int i = 1; i <= tot; ++i)
        vec.push_back(l[i]);
    cin >> q;
    for (int i = 1; i <= q; ++i)
    {
        cin >> qry[i];
        vec.push_back(qry[i]);
    }
    sort(all(vec));
    vec.erase(unique(all(vec)), vec.end());
    int m = vec.size() + 10;
    build(1, 1, m);
    for (int i = 1; i <= tot; ++i)
        change(1, 1, m, find(l[i]), r[i]);
    for (int i = 1; i <= q; ++i)
    {
        int x = find(qry[i]);
        int mx = query(1, 1, m, 1, x).mx;
        cout << max(mx, qry[i]) << " ";
    }
    cout << endl;
}