拿小小号打的DIV3，中间看了会儿b站摸鱼，结果尼玛最后几点钟G没写完。。。
A. Robin Helps
模拟题
int T, n, k;signed main(void) {for (read(T); T; T--) {read(n), read(k); int ans = 0; ll sum = 0;for (int i = 1; i <= n; i++) {int x; read(x);if (x >= k) sum += x;if (x == 0) {if (sum) sum--, ans++;}}writeln(ans);}//fwrite(pf, 1, o1 - pf, stdout);return 0;
}
B. Robin Hood and the Major Oak
根据mod4分类
int T, n, k;signed main(void) {for (read(T); T; T--) {read(n), read(k);if (k % 4 == 0 || (k % 4 == 1 && n % 2 == 0) || (n % 2 == 1 && k % 4 == 3)) puts("YES");else puts("NO"); }//fwrite(pf, 1, o1 - pf, stdout);return 0;
}C. Robin Hood in Town
二分一下\(\Delta\)即可。
const int N = 2e5 + 5;
int T, n, k, a[N]; ll sum;inline bool check(ll x) {int ret = 0;for (int i = 2; i <= n; i++)if (2ll * n * a[i] < sum + x) ++ret;if (ret * 2 > n) return true;return false;
}signed main(void) {for (read(T); T; T--) {read(n); int mx = 0; sum = 0;for (int i = 1; i <= n; i++)read(a[i]), chkmax(mx, a[i]), sum += a[i];if (n == 1 || n == 2) { puts("-1"); continue; }sort(a + 1, a + n + 1, greater <int> ());ll l = 0, r = (ll) 1e18, ans = (ll) 1e18;while (l <= r) {ll mid = l + r >> 1;if (check(mid)) ans = mid, r = mid - 1;else l = mid + 1;}writeln(ans);}//fwrite(pf, 1, o1 - pf, stdout);return 0;
}D. Robert Hood and Mrs Hood
用set模拟题意即可。
const int N = 1e5 + 10;
int n, d, k;
vector<int> l[N], r[N];void slv() {cin >> n >> d >> k;for(int i = 1; i <= n; i++) l[i].clear(), r[i].clear();for(int i = 1; i <= k; i++) {int x, y;cin >> x >> y;l[x].push_back(i), r[y].push_back(i);}set<int> st;int ans1 = 0, ans2 = k + 1, ip1, ip2, pos = 1;for(int i = 1; i <= d; i++) {for(int ip : l[i]) st.insert(ip);}ans1 = ans2 = st.size(), ip1 = ip2 = 1;// cerr<<" pos= "<<1<<" siz="<<st.size()<<'\n';while(pos + 1 + d - 1 <= n) {for(int i : l[pos + 1 + d - 1]) st.insert(i);for(int i : r[pos]) st.erase(i);// cerr<<" pos= "<<pos+1<<" siz="<<st.size()<<'\n';if(ans1 < st.size()) ans1 = st.size(), ip1 = pos + 1;if(ans2 > st.size()) ans2 = st.size(), ip2 = pos + 1;pos++;}cout << ip1 << ' ' << ip2 << '\n';return;
}
E. Rendez-vous de Marian et Robin
跑两遍分层图最短路，枚举交点取最小的最大值即可。
struct Edge {int v, w;Edge(int v = 0, int w = 0) : v(v), w(w) {}
};const int N = 2e5 + 5;
int T, n, m, h, a[N];
vector <Edge> G[N];ll dis[N][2], f[N], g[N]; bool inq[N][2];
inline void DijkstraI(int s) {for (int i = 1; i <= n; i++) dis[i][0] = dis[i][1] = (ll) 1e18;priority_queue <piii, vector<piii>, greater<piii> > q;dis[s][0] = 0; q.push(Mpp(dis[s][0], s, 0)); Ms(inq, 0);if (a[s]) dis[s][1] = 0, q.push(Mpp(dis[s][1], s, 1));while (!q.empty()) {int u = q.top().second.first, id = q.top().second.second; q.pop();if (inq[u][id]) continue; inq[u][id] = true;for (auto x : G[u]) {int v = x.v, w = x.w;int e = id ? w / 2 : w;if (dis[v][id] > dis[u][id] + e) {dis[v][id] = dis[u][id] + e;if (!inq[v][id]) q.push(Mpp(dis[v][id], v, id));}if (a[v] && id == 0 && dis[v][1 - id] > dis[u][id] + e) {dis[v][1 - id] = dis[u][id] + e;if (!inq[v][1 - id]) q.push(Mpp(dis[v][1 - id], v, 1 - id));}}}for (int i = 1; i <= n; i++) f[i] = min(dis[i][1], dis[i][0]);
}inline void DijkstraII(int s) {for (int i = 1; i <= n; i++) dis[i][0] = dis[i][1] = (ll) 1e18;priority_queue <piii, vector<piii>, greater<piii> > q;dis[s][0] = 0; q.push(Mpp(dis[s][0], s, 0)); Ms(inq, 0);if (a[s]) dis[s][1] = 0, q.push(Mpp(dis[s][1], s, 1));while (!q.empty()) {int u = q.top().second.first, id = q.top().second.second; q.pop();if (inq[u][id]) continue; inq[u][id] = true;for (auto x : G[u]) {int v = x.v, w = x.w;int e = id ? w / 2 : w;if (dis[v][id] > dis[u][id] + e) {dis[v][id] = dis[u][id] + e;if (!inq[v][id]) q.push(Mpp(dis[v][id], v, id));}if (a[v] && id == 0 && dis[v][1 - id] > dis[u][id] + e) {dis[v][1 - id] = dis[u][id] + e;if (!inq[v][1 - id]) q.push(Mpp(dis[v][1 - id], v, 1 - id));}}}for (int i = 1; i <= n; i++) g[i] = min(dis[i][1], dis[i][0]);
}signed main(void) {for (read(T); T; T--) {read(n), read(m), read(h);for (int i = 1; i <= n; i++) G[i].clear(), a[i] = 0;for (int i = 1, x; i <= h; i++) read(x), a[x] = 1;for (int i = 1, u, v, w; i <= m; i++) {read(u), read(v), read(w);G[u].push_back(Edge(v, w));G[v].push_back(Edge(u, w));}DijkstraI(1); DijkstraII(n); ll ans = (ll) 1e18;for (int i = 1; i <= n; i++) {if (f[i] == (ll) 1e18 || g[i] == (ll) 1e18) continue;chkmin(ans, max(f[i], g[i]));}writeln(ans == (ll) 1e18 ? -1 : ans);}//fwrite(pf, 1, o1 - pf, stdout);return 0;
}
某个傻逼wa两次的原因：
F. Sheriff's Defense
简单树形dp
const int N = 2e5 + 5;
int T, n, c, a[N];
vector <int> G[N];ll dp[N][2];
inline void dfs(int u, int f) {dp[u][1] = a[u];for (auto v : G[u]) {if (v == f) continue;dfs(v, u);dp[u][0] += max(dp[v][0], dp[v][1]);dp[u][1] += max(dp[v][0], dp[v][1] - 2 * c);}
}signed main(void) {for (read(T); T; T--) {read(n), read(c);for (int i = 1; i <= n; i++)read(a[i]), dp[i][0] = dp[i][1] = 0, G[i].clear();for (int i = 1, u, v; i < n; i++) {read(u), read(v);G[u].push_back(v);G[v].push_back(u);}dfs(1, 0);writeln(max(dp[1][0], dp[1][1]));}//fwrite(pf, 1, o1 - pf, stdout);return 0;
}
G. Milky Days
傻逼双端队列模拟题，就是细节有点多。代码就不放了，懂得都懂。
H. Robin Hood Archery
本质是询问一个区间里所有出现的数是不是出现偶数次，可以一眼莫队，也可以用随机数加的异或前缀和。
inline void slv(void) {cin >> n >> m;siz = sqrt(n);map<int, int> mp;int cnt = 0;for(int i = 1; i <= n; i++) {cin >> b[i];if(!mp[b[i]]) mp[b[i]] = ++cnt;b[i] = mp[b[i]];}s.clear(cnt);for(int i = 1; i <= m; i++) {cin >> a[i].l >> a[i].r;a[i].id = i;a[i].pos = (a[i].l - 1) / siz + 1;}sort(a + 1, a + 1 + m);int l = 1, r = 0;ll ans = 0;for(int i = 1; i <= m; i++) {while(l > a[i].l) s.ins(b[--l]);while(r < a[i].r) s.ins(b[++r]);while(l < a[i].l) s.del(b[l++]);while(r > a[i].r) s.del(b[r--]);res[a[i].id] = s.chk();}for(int i = 1; i <= m; i++)if(res[i]) cout << "YES\n";else cout << "NO\n";return;
}
