C. Robot Factory

用前 \(k\) 小的头和前 \(k\) 大的身体按递增顺序做匹配

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)

using namespace std;

int main() {
    int n, m, k;
    cin >> n >> m >> k;
    
    vector<int> h(n), b(m);
    rep(i, n) cin >> h[i];
    rep(i, m) cin >> b[i];
    
    sort(h.begin(), h.end());
    sort(b.begin(), b.end());
    
    vector<int> kh(h.begin(), h.begin()+k);
    vector<int> kb(b.end()-k, b.end());
    
    rep(i, k) if (kh[i] > kb[i]) {
        puts("No");
        return 0;
    }
    
    puts("Yes");

    return 0;
}

D. Robot Customize

一开始不妨将所有零件安装在身体上，然后对于所有 \(H_i > B_i\) 的零件做 \(01\) 背包（价值为 \(H_i-B_i\)），为了满足头部重量不超过身体重量，将背包容量设置为 \(\frac{1}{2}\sum W_i\)

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)

using namespace std;
using ll = long long;
using P = pair<int, int>;

inline void chmax(ll& a, ll b) { if (a < b) a = b; }

int main() {
    int n;
    cin >> n;
    
    int W = 0; ll sumb = 0;
    vector<P> ps;
    rep(i, n) {
        int w, h, b;
        cin >> w >> h >> b;
        W += w;
        sumb += b;
        if (h > b) ps.emplace_back(w, h-b);
    }
    
    const ll INF = 1e18;
    n = ps.size();
    W /= 2;
    vector<ll> dp(W+1);
    
    rep(i, n) {
        vector<ll> old(W+1, -INF);
        swap(dp, old);
        
        auto [w, v] = ps[i];
        rep(j, W+1) {
            chmax(dp[j], old[j]);
            if (j+w <= W) chmax(dp[j+w], old[j]+v);
        }
    }
    
    ll ans = dp[W]+sumb;
    cout << ans << '\n';
    
    return 0;
}

E. Reflection on Grid

因为更换镜子时可以任意调整朝向，所以所有需要二次访问的解都能无损地改造成一次访问的解。因此只要考虑简单路径的情况即可。以当前位置坐标和当前朝向作为状态，用 \(01\text{bfs}\) 求解。

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)

using namespace std;
using ll = long long;
using P = pair<int, int>;

const int di[] = {-1, 0, 1, 0};
const int dj[] = {0, -1, 0, 1};

void solve() {
    int h, w;
    cin >> h >> w;
    
    vector<string> s(h);
    rep(i, h) cin >> s[i];
    
    const int INF = 1001001001;
    vector dist(h, vector(w, vector<int>(4, INF)));
    deque<tuple<int, int, int, int>> q;
    auto push = [&](int i, int j, int v, int d, int cost) {
        d += cost;
        if (dist[i][j][v] <= d) return;
        dist[i][j][v] = d;
        if (cost == 0) q.emplace_front(d, i, j, v);
        else q.emplace_back(d, i, j, v);
    };
    push(0, 0, 3, 0, 0);
    
    int ans = INF;
    while (q.size()) {
        auto [d, i, j, v] = q.front(); q.pop_front();
        if (dist[i][j][v] != d) continue;
        int ov = v;
        if (s[i][j] == 'B') ov ^= 1;
        if (s[i][j] == 'C') ov ^= 3;
        rep(u, 4) {
            int ni = i+di[u], nj = j+dj[u];
            int cost = ov==u?0:1;
            if (ni == h-1 and nj == w) {
                ans = min(ans, d+cost);
            }
            if (ni < 0 or ni >= h or nj < 0 or nj >= w) continue;
            push(ni, nj, u, d, cost);
        }
    }
    
    cout << ans << '\n';
}

int main() {
    int t;
    cin >> t;
    
    while (t--) solve();
    
    return 0;
}

F. Almost Sorted 2

对于一个数 \(X\)，只能插进 \([x-D, X-1]\) 之间的空隙，或者插在末尾
比较简单的组合计数，假设在插入 \(X\) 之前已经有 \(K\) 个位于 \([x-D, X-1]\) 之间的数了，且 \(X\) 有 \(m\) 个，那么这 \(m\) 个 \(X\) 的插法数就是 \(\binom{m+k-1}{m}\)。我们利用乘法原理，就能算出最终的答案。
背后的算法思想是插入 \(\text{dp}\)

双倍经验：P6522 (注：这个不需要去重)

代码实现 
#include <bits/stdc++.h>
#include <atcoder/all>
using namespace atcoder;
#define rep(i, n) for (int i = 0; i < (n); ++i)

using namespace std;
using mint = modint998244353;

struct modinv {
  int n; vector<mint> d;
  modinv(): n(2), d({0,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(-d[mint::mod()%n]*(mint::mod()/n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} invs;
struct modfact {
  int n; vector<mint> d;
  modfact(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*n), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} facts;
struct modfactinv {
  int n; vector<mint> d;
  modfactinv(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*invs(n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} ifacts;
mint comb(int n, int k) {
  if (n < k || k < 0) return 0;
  return facts(n)*ifacts(k)*ifacts(n-k);
}

int main() {
    int n, d;
    cin >> n >> d;
    
    const int M = 1e6+5;
    vector<int> c(M);
    rep(i, n) {
        int a;
        cin >> a;
        c[a]++;
    }
    
    mint ans = 1;
    int k = 1;
    rep(i, M) {
        ans *= comb(c[i]+k-1, c[i]);
        k += c[i];
        if (i >= d) k -= c[i-d];
    }
    
    cout << ans.val() << '\n';
    
    return 0;
}

G. One Time Swap 2

参考 StarSilk