拿小小號打的DIV3，中間看了會兒b站摸魚，結果尼瑪最後幾點鐘G沒寫完。。。
A. Robin Helps
模擬題
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
根據mod4分類
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模擬題意即可。
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
跑兩遍分層圖最短路，列舉交點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;
}
某個傻逼wa兩次的原因：
F. Sheriff's Defense
簡單樹形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
傻逼雙端佇列模擬題，就是細節有點多。程式碼就不放了，懂得都懂。
H. Robin Hood Archery
本質是詢問一個區間裡所有出現的數是不是出現偶數次，可以一眼莫隊，也可以用隨機數加的異或字首和。
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;
}