ABC434

C. Flapping Takahashi

维护当前时刻 \(t\) 可达的高度区间 \([\text{lh}, \text{rh}]\) 。从时刻 \(t_{i-1}\) 到 \(t_i\) 区间扩散为 \([\text{lh}-\Delta t, \text{rh} + \Delta t]\) 。 然后对区间 \([l, u]\) 取交集。

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;void solve() {int n, h;cin >> n >> h;int now = 0, lh = h, rh = h;bool ans = true;rep(i, n) {int t, l, u;cin >> t >> l >> u;lh -= t-now; rh += t-now;now = t;lh = max(lh, l);rh = min(rh, u);if (lh > rh) ans = false;}if (ans) puts("Yes");else puts("No");
}int main() {int t;cin >> t;while (t--) solve();return 0;
}

D. Clouds

二维差分
显然只有被 \(1\) 朵云覆盖的格子才会影响答案

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;const int M = 2000;
int sky[M+1][M+1];
int one[M+1][M+1];struct Cloud { int u, d, l, r; };int main() {int n;cin >> n;vector<Cloud> clouds;rep(i, n) {int u, d, l, r;cin >> u >> d >> l >> r;--u; --l;sky[u][l]++;sky[u][r]--;sky[d][l]--;sky[d][r]++;clouds.emplace_back(u, d, l, r);}rep(i, M)rep(j, M) sky[i+1][j] += sky[i][j];rep(i, M)rep(j, M) sky[i][j+1] += sky[i][j];rep(i, M)rep(j, M) one[i+1][j+1] = sky[i][j]==1;rep(i, M)rep(j, M+1) one[i+1][j] += one[i][j];rep(i, M+1)rep(j, M) one[i][j+1] += one[i][j];int base = 0;rep(i, M)rep(j, M) if (sky[i][j] == 0) base++;rep(i, n) {auto [u, d, l, r] = clouds[i];int dif = one[d][r] - one[d][l] - one[u][r] + one[u][l];int ans = base + dif;cout << ans << '\n';}return 0;
}

E. Distribute Bunnies

一种做法是二分图最大匹配
另一种做法是单独考虑每个无向图连通分量，为其定向：

• 边数 \(<\) 点数，树
• 边数 \(=\) 点数，基环树
• 边数 \(>\) 点数，把重边删掉就是等于的情况

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;
using P = pair<int, int>;// Coodinate Compression
template<typename T=int>
struct CC {bool initialized;vector<T> xs;CC(): initialized(false) {}void add(T x) { xs.push_back(x);}void init() {sort(xs.begin(), xs.end());xs.erase(unique(xs.begin(),xs.end()),xs.end());initialized = true;}int operator()(T x) {if (!initialized) init();return upper_bound(xs.begin(), xs.end(), x) - xs.begin() - 1;}T operator[](int i) {if (!initialized) init();return xs[i];}int size() {if (!initialized) init();return xs.size();}
};int main() {int m;cin >> m;vector<P> bunnies;CC cc;rep(i, m) {int x, r;cin >> x >> r;bunnies.emplace_back(x-r, x+r);cc.add(x-r); cc.add(x+r);}int n = cc.size();vector<vector<int>> to(n);rep(i, m) {auto [a, b] = bunnies[i];a = cc(a); b = cc(b);to[a].push_back(b);to[b].push_back(a);}int ans = 0;vector<bool> used(n);rep(i, n) if (!used[i]) {int edge = 0, vertex = 0;auto dfs = [&](auto& f, int v) -> void {if (used[v]) return;vertex++; edge += to[v].size();used[v] = true;for (int u : to[v]) f(f, u);};dfs(dfs, i);edge /= 2;ans += min(vertex, edge);}cout << ans << '\n';return 0;
}

F. Concat (2nd)

按 \(S_i + S_j < S_j + S_i\) 排序
显然直接排显然超时
有两种优化方法：

• 先 \(\text{shuffle}\) 一遍再排
• 用 \(z\) 算法优化比较函数

代码实现1

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;void solve() {int n;cin >> n;vector<string> s(n);rep(i, n) cin >> s[i];random_shuffle(s.begin(), s.end());sort(s.begin(), s.end(), [&](const string& a, const string& b) {return a+b < b+a;});auto cat = [&]() {string res;for (string& t : s) res += t;return res;};rep(i, n-1) {if (s[i]+s[i+1] == s[i+1]+s[i]) {cout << cat() << '\n';return;}}swap(s[n-2], s[n-1]);string ans = cat();swap(s[n-2], s[n-1]);if (n >= 3) {swap(s[n-2], s[n-3]);ans = min(ans, cat());}cout << ans << '\n';
}int main() {int t;cin >> t;while (t--) solve();return 0;
}

代码实现2

#include <bits/stdc++.h>
#include <atcoder/all>
using namespace atcoder;
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;void solve() {int n;cin >> n;vector<string> s(n);rep(i, n) cin >> s[i];vector<vector<int>> z(n);rep(i, n) z[i] = z_algorithm(s[i]);auto comp = [&](int i, int j) {int w = s[i].size();if (s[i] != s[j].substr(0, w)) return (s[i] < s[j])?-1:1;int r = s[j].size()-w;if (r > 0 and z[j][w] < r) {int k = z[j][w];return (s[j][k] < s[j][w+k])?-1:1;}return (s[j].substr(r) < s[i])?-1:1;};vector<int> p(n);rep(i, n) p[i] = i;sort(p.begin(), p.end(), [&](int i, int j) {if (s[i].size() > s[j].size()) return comp(j, i) == 1;return comp(i, j) == -1;});{auto old = s;rep(i, n) s[i] = old[p[i]];}auto cat = [&]() {string res;for (string& t : s) res += t;return res;};rep(i, n-1) {if (s[i]+s[i+1] == s[i+1]+s[i]) {cout << cat() << '\n';return;}}swap(s[n-2], s[n-1]);string ans = cat();swap(s[n-2], s[n-1]);if (n >= 3) {swap(s[n-2], s[n-3]);ans = min(ans, cat());}cout << ans << '\n';
}int main() {int t;cin >> t;while (t--) solve();return 0;
}