The 2022 ICPC Asia Shenyang Regional Contest

C. Clamped Sequence

因为\(n\)的范围不大，并且可以猜到\(l,r\)中应该至少有一个在\(a_i，a_i-1,a_i+1\)上。所以直接暴力枚举\(l\)或\(r\)然后暴力的计算一下

#include <bits/stdc++.h>

using namespace std;

#define int long long

using vi = vector<int>;


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, d, res = 0;
    cin >> n >> d;
    vector<int> a(n), b;
    for (int &i: a)
        cin >> i, b.push_back(i), b.push_back(i - 1), b.push_back(i + 1);
    sort(b.begin(), b.end());
    b.resize(unique(b.begin(), b.end()) - b.begin());
    for (int r; int l: b) {
        r = l + d;
        auto c = a;
        for (auto &i: c) {
            if (i < l) i = l;
            if (i > r) i = r;
        }
        int cnt = 0;
        for (int i = 1; i < n; i++)
            cnt += abs(c[i] - c[i - 1]);
        res = max(res, cnt);
    }
    for (int l; int r: b) {
        l = r - d;
        auto c = a;
        for (auto &i: c) {
            if (i < l) i = l;
            if (i > r) i = r;
        }
        int cnt = 0;
        for (int i = 1; i < n; i++)
            cnt += abs(c[i] - c[i - 1]);
        res = max(res, cnt);
    }
    cout << res << "\n";
    return 0;
}

D. DRX vs. T1

签到

#include <bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    string s;
    cin >> s;
    map<char,int> cnt;
    for( auto i : s )
        cnt[i] ++;
    if( cnt['T'] >= 3 ) cout << "T1\n";
    else if( cnt['D'] >= 3) cout << "DRX\n";
    return 0;
}

F. Half Mixed

首先整个矩形的子矩形数量有\(\frac{n(n+2)}{2}\times\frac{m(m+1)}{2}\)个，如果这个数一奇数则一定无解。

这样的话，我们考虑构造某一行或一列然后复制即可。

假设我们构造$1\times m $ 的矩形，结果一定是若干段相同且连续的子段，设每一段长度为\(l_i\)，则这一段的纯净矩阵数量为\(\frac{l_i(l_i+1)}{2}\)。

所以要满足以下的条件

\[\sum l_i = m \and \sum\frac{l_i(l_i+1)}{2}=\frac{m(m+1)}{4} \]

首先考虑要满足第一个条件，则纯净矩阵至少有\(m\)个，那么每个子段除去本身外的纯净矩阵就剩下了\(\sum \frac{l_i(l_i-1)}{2}\)

所以即在满足

\[\sum\frac{l_i(l_i-1)}{2}=\frac{m(m+1)}{4}-m \]

的情况下，每次\(l_i\)取最大值即可

#include <bits/stdc++.h>

using namespace std;

#define int long long

using vi = vector<int>;

void solve() {
    int n, m;
    cin >> n >> m;
    if ((n * (n + 1) * m * (m + 1) / 4) & 1) {
        cout << "No\n";
        return;
    }
    cout << "Yes\n";
    int t = n, T = n * (n + 1) / 2;
    if (T & 1) t = m, T = m * (m + 1) / 2;
    T = T / 2 - t;
    auto f = [&t, &T]() {
        int l = 1, r = t, res = -1;
        for (int mid; l <= r;) {
            mid = (l + r) / 2;
            if (mid * (mid - 1) / 2 <= T) res = mid, l = mid + 1;
            else r = mid - 1;
        }
        t -= res, T -= res * (res - 1) / 2;
        return res;
    };
    vector<int> res;
    for (int x = 0, y; t; x ^= 1) {
        y = f();
        for (; y; y--) res.push_back(x);
    }
    if (res.size() == m) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++)
                cout << res[j] << " ";
            cout << "\n";
        }
    } else {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cout << res[i] << " ";
            }
            cout << "\n";
        }
    }
    return;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int TC;
    for (cin >> TC; TC; TC--)
        solve();
    return 0;
}

L. Tavern Chess

直接根据题意模拟，模拟的形式就是用 dfs 去搜索，在搜索的过程中计算出到当前状态的概率。

When a team takes the attack, the leftmost minion with the mininum number of taking attacks from the team attacks one of the alive minios from the other team uniformyly ad random, and then the other team takes the attack.

这句话，在赛时卡了很久，我认为其正确的意思应该是：当一个队伍攻击时，发起攻击次数最少且最靠左的士兵会等概率的随机攻击另一方一个存活的士兵，在攻击后交换攻击。

#include <bits/stdc++.h>

using namespace std;

using vi = vector<int>;
using db = long double;

int n, m;
vi atk0, atk1;
db A = 0, B = 0, T = 0;

void dfs(vi hp0, vi hp1, vi cnt0, vi cnt1, int t, db p, int dep) {

    db alive0 = 0, alive1 = 0; // 存活人数
    for (auto i: hp0) alive0 += (i > 0);
    for (auto i: hp1) alive1 += (i > 0);


    if (alive0 == 0 or alive1 == 0) { // 游戏结束
        if (alive0 == alive1) T += p;
        else if (alive0 == 0) B += p;
        else A += p;
        return;
    }


    if (t == 0) { // Alice attack
        int id = -1;
        for (int i = 0; i < n; i++) {
            if (hp0[i] <= 0) continue;
            if (id == -1 or cnt0[id] > cnt0[i]) id = i;
        }
        cnt0[id]++;

        for (int i = 0; i < m; i++) {
            if (hp1[i] <= 0) continue;
            hp0[id] -= atk1[i], hp1[i] -= atk0[id];
            dfs(hp0, hp1, cnt0, cnt1, t ^ 1, p / alive1, dep + 1);
            hp0[id] += atk1[i], hp1[i] += atk0[id];
        }
    } else { // Bob attack
        int id = -1;
        for (int i = 0; i < m; i++) {
            if (hp1[i] <= 0) continue;
            if (id == -1 or cnt1[id] > cnt1[i]) id = i;
        }
        cnt1[id]++;
        for (int i = 0; i < n; i++) {
            if (hp0[i] <= 0) continue;
            hp0[i] -= atk1[id], hp1[id] -= atk0[i];
            dfs(hp0, hp1, cnt0, cnt1, t ^ 1, p / alive0, dep + 1);
            hp0[i] += atk1[id], hp1[id] += atk0[i];
        }
    }
    return;
}


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    cin >> n >> m;
    atk0 = vi(n), atk1 = vi(m);
    for (auto &i: atk0) cin >> i;
    for (auto &i: atk1) cin >> i;
    if (n > m) dfs(atk0, atk1, vi(n), vi(m), 0, 1.0, 0);
    else if (m > n) dfs(atk0, atk1, vi(n), vi(m), 1, 1.0, 0);
    else dfs(atk0, atk1, vi(n), vi(m), 0, 0.5, 0), dfs(atk0, atk1, vi(n), vi(m), 1, 0.5, 0);

    cout << fixed << setprecision(10) << A << "\n" << B << "\n" << T << "\n";
    return 0;
}