『模拟赛』NOIP2024加赛5

Rank

反向挂分大王

A. 暴力操作（opt）

签，但是没有人签。

都想到了二分和更新 c 值，但是 c 多多少少没更到最优。

首先还是调和级数预处理，倒序取 min。然后考虑到超过 \(m\) 的也有可能产生更小的代价，因此 \(\mathcal{O(n)}\) 枚举一遍找到最小的 \(j\) 使 \(i\times j\gt m\)，全部赋到 \(c_{m+1}\) 上，然后再倒序取一遍 min 就能更好的更新。

check 有 \(\mathcal{O(n)}\) 的双指针写法，我直接没改赛时的 \(\mathcal{O(n\log n)}\) 的二分，所以复杂度是 \(\mathcal{O(n\ln n+n\log^2n)}\) 的。

点击查看代码

#include<bits/stdc++.h>
#define fo(x, y, z) for(int (x) = (y); (x) <= (z); (x)++)
#define fu(x, y, z) for(int (x) = (y); (x) >= (z); (x)--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define lx ll
inline lx qr()
{char ch = getchar(); lx x = 0, f = 1;for(; ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);return x * f;
}
#undef lx
#define qr qr()
#define pii pair<int, int>
#define ppp pair<pii, pii>
#define fi first
#define se second
#define M_P(x, y) make_pair(x, y)
#define P_B(x) push_back(x)
const int Ratio = 0;
const int N = 5e5 + 5;
const int mod = 1e9 + 7;
ll n, m, k;
ll a[N], c[N];
namespace Wisadel
{bool Wck(int x){ll ned = 0;fo(i, 1, n / 2 + 1) if(a[i] > x){int l = 2, r = a[i] + 1;while(l < r){int mid = (l + r) >> 1;if(a[i] / mid <= x) r = mid;else l = mid + 1;}ned += c[r];if(ned > k) return 0;}return 1;}short main(){freopen("opt.in", "r", stdin), freopen("opt.out", "w", stdout);n = qr, m = qr, k = qr;fo(i, 1, n) a[i] = qr;fo(i, 1, m) c[i] = qr;fo(i, 2, m) fo(j, 1, min(m / i, 1ll * i))c[i * j] = min(c[i * j], c[i] + c[j]);fu(i, m - 1, 1) c[i] = min(c[i], c[i + 1]);c[m + 1] = 1e9;fo(i, 2, m){int zc = m / i + 1;c[m + 1] = min(c[m + 1], c[i] + c[zc]);}fu(i, m, 1) c[i] = min(c[i], c[i + 1]);sort(a + 1, a + 1 + n);int l = 0, r = a[n / 2 + 1];while(l < r){int mid = (l + r) >> 1;if(Wck(mid)) r = mid;else l = mid + 1;}printf("%d\n", r);return Ratio;}
}
signed main(){return Wisadel::main();}
// Now there's only one thing I can do
// Fight until the end like I promised to

B. 异或连通（xor）

比较套路。

主要需要想到对询问建 trie 而不是边，那么容易发现一条边只会在 \(\log V\) 个子树中出现。按线段树分治的思想，直接在 trie 树上分治，找到对于每一个询问合法的线段存到点上，然后对 trie 树做一遍 dfs，过程中做启发式合并，回溯时撤销即可。

注意并查集不要路径压缩！复杂度 \(\mathcal{O(n\log n\log V)}\)。

点击查看代码

#include<bits/stdc++.h>
#define fo(x, y, z) for(int (x) = (y); (x) <= (z); (x)++)
#define fu(x, y, z) for(int (x) = (y); (x) >= (z); (x)--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define lx ll
inline lx qr()
{char ch = getchar(); lx x = 0, f = 1;for(; ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);return x * f;
}
#undef lx
#define qr qr()
#define pii pair<int, int>
#define fi first
#define se second
#define M_P(x, y) make_pair(x, y)
#define P_B(x) push_back(x)
const int Ratio = 0;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
int n, m, k, q;
struct rmm{int u, v, w;} e[N];
int que[N], pre[N];
namespace Wisadel
{ll fx[N], siz[N];int Wfind(int x){if(x == fx[x]) return x;return Wfind(fx[x]);}void Wmerge(int u, int v){u = Wfind(u), v = Wfind(v);if(u == v) return ;if(siz[u] < siz[v])fx[u] = v, siz[v] += siz[u];else fx[v] = u, siz[u] += siz[v];}int t[N * 20][2], tot;vector<pii> d[N * 20];vector<int> v[N * 20];ll ans[N * 20];void Wins(int x, int id){int now = 0;fu(i, 30, 0){int to = (x >> i) & 1;if(!t[now][to]) t[now][to] = ++tot;now = t[now][to];}pre[id] = now;}void Wupd(int x, int id){int now = 0;fu(i, 30, 0){int to = (x >> i) & 1, pd = (k >> i) & 1;if(pd){if(t[now][to]) v[t[now][to]].P_B(id);now = t[now][to ^ 1];}else now = t[now][to];if(!now) return ;}}void Wdfs(int now, ll sum){ans[now] = sum;for(int i : v[now]){int _ = Wfind(e[i].u), __ = Wfind(e[i].v);if(_ == __) continue;if(siz[_] < siz[__]) swap(_, __);ans[now] += siz[_] * siz[__];fx[__] = _, siz[_] += siz[__];d[now].P_B(M_P(_, __));}if(t[now][0]) Wdfs(t[now][0], ans[now]);if(t[now][1]) Wdfs(t[now][1], ans[now]);fu(i, d[now].size() - 1, 0){pii zc = d[now][i];fx[zc.se] = zc.se;siz[zc.fi] -= siz[zc.se];}}short main(){freopen("xor.in", "r", stdin), freopen("xor.out", "w", stdout);n = qr, m = qr, q = qr, k = qr;fo(i, 1, n) fx[i] = i, siz[i] = 1;fo(i, 1, m) e[i].u = qr, e[i].v = qr, e[i].w = qr;fo(i, 1, q) que[i] = qr, Wins(que[i], i);fo(i, 1, m) Wupd(e[i].w, i);Wdfs(0, 0);fo(i, 1, q) printf("%lld\n", ans[pre[i]]);return Ratio;}
}
signed main(){return Wisadel::main();}
// Now there's only one thing I can do
// Fight until the end like I promised to

C. 诡异键盘（keyboard）

huge：你们上届学长只有三个改出来了

放最后改了，懂了大概思路。

赛时纯唐纯暴力做法打的 20pts 拿到了 70pts。

先做删串需要的同余最短路，然后建 trie 做 dp，具体不太会。

D. 民主投票（election）

好像是真正的签，不过为什么我需要卡常才能过？

首先需要求得树上最多票数最少 \(k\) 是多少，这个容易想到二分去做，设 \(f_i\) 为子树 \(i\) 在某个票数限制下需要向上贡献的票数，判断可行与否只需根据 \(f1=0\)。然后对于子树大小减一大于 \(k\) 点一定可行，小于 \(k\) 的一定不行，这两点都很显然。

关键在处理子树大小减一等于 \(k\) 的点。办法是以 \(k-1\) 为限制再 dp 一次。此时只有当前点的限制为 \(k\)，则若 \(f_x=1\)，那么解除限制后才满足 \(f_x=0\)。只有 \(x\) 到根路径上的点的 \(f\) 都为 0 且 \(f_1=1\) 时才合法，因为只有此时 \(f_x\) 限制加一才能解除 \(f_1\) 的限制。

时间复杂度 \(\mathcal{O(n\log n)}\)，不知道为啥要卡常。

点击查看代码

// ubsan: undefined
// accoders
#include<bits/stdc++.h>
#define fo(x, y, z) for(int (x) = (y); (x) <= (z); (x)++)
#define fu(x, y, z) for(int (x) = (y); (x) >= (z); (x)--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define lx ll
inline lx qr()
{char ch = getchar(); lx x = 0, f = 1;for(; ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);return x * f;
}
#undef lx
#define qr qr()
#define pii pair<int, int>
#define ppp pair<pii, pii>
#define fi first
#define se second
#define M_P(x, y) make_pair(x, y)
#define P_B(x) push_back(x)
const int Ratio = 0;
const int N = 2e6 + 5;
const int mod = 1e9 + 7;
int n;
int hh[N], to[N], ne[N], cnt;
int siz[N], f[N];
bool ok[N];
namespace Wisadel
{inline void Wadd(int u, int v){to[++cnt] = v;ne[cnt] = hh[u];hh[u] = cnt;}inline void Wdfs(int u, int fa){siz[u] = 1;for(int i = hh[u]; i != -1; i = ne[i]){int v = to[i];Wdfs(v, u);siz[u] += siz[v];}}inline void Wck(int u, int tar){f[u] = 0;for(int i = hh[u]; i != -1; i = ne[i]){int v = to[i];Wck(v, tar);f[u] += f[v] + 1;}if(f[u] <= tar) f[u] = 0;else f[u] -= tar;}inline void Wsol(int u){if((u == 1 && f[u] != 1) || !f[u]) return ;if(f[u] == 1) ok[u] = 1;for(int i = hh[u]; i != -1; i = ne[i]){int v = to[i];Wsol(v);}}short main(){freopen("election.in", "r", stdin), freopen("election.out", "w", stdout);int T = qr;while(T--){n = qr;cnt = 0;fill(hh + 1, hh + 1 + n, -1);fo(i, 2, n){int x = qr;Wadd(x, i);}Wdfs(1, 0);int l = 1, r = n, ke = 0;while(l <= r){int mid = (l + r) >> 1;Wck(1, mid);if(f[1] == 0) r = mid - 1, ke = mid;else l = mid + 1;}memset(ok+1, 0, sizeof(bool)*n);Wck(1, ke - 1);Wsol(1);fo(i, 1, n){int zc = siz[i] - 1;if(zc > ke) putchar('1');else if(zc < ke) putchar('0');else printf("%d", ok[i]);}puts("");}return Ratio;}
}
signed main(){return Wisadel::main();}
// Now there's only one thing I can do
// Fight until the end like I promised to

邀请了 jijidawang 为特邀嘉宾帮助卡常，直接将一个 7000ms+ 的代码通过各种科技卡到了 accoder 上 1380ms。

jjdw 科技代码

// ubsan: undefined
// accoders
#include<bits/stdc++.h>
#define int unsigned
#define fo(x, y, z) for(int (x) = (y); (x) <= (z); (x)++)
#define fu(x, y, z) for(int (x) = (y); (x) >= (z); (x)--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define lx ll
inline lx qr()
{char ch = getchar(); lx x = 0, f = 1;for(; ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);return x * f;
}
#undef lx
#define qr qr()
#define pii pair<int, int>
#define ppp pair<pii, pii>
#define fi first
#define se second
#define M_P(x, y) make_pair(x, y)
#define P_B(x) push_back(x)
constexpr int Ratio = 0;
constexpr int N = 1e6 + 5;
const int mod = 1e9 + 7;
int n;
int siz[N], f[N], qwq[N];
int ok[N];
namespace Wisadel
{vector<int> g[N];inline void Wadd(int u, int v){g[u].emplace_back(v);}inline void Wck(int u, signed tar){memset(f+1,0,sizeof(int)*n);#pragma GCC unroll 16for (int i=n; i>=2; i--){f[i] = max<signed>((signed)f[i] - tar, 0);f[qwq[i]] += f[i] + 1;}f[1] = max<signed>((signed)f[1] - tar, 0);}inline void Wsol(int u){if((u == 1 && f[u] != 1) || !f[u]) return ;if(f[u] == 1) ok[u] = 1;#pragma GCC unroll 2for (int v : g[u])Wsol(v);}inline __int32_t main(){freopen("election.in", "r", stdin), freopen("election.out", "w", stdout);int T = qr;while(T--){n = qr;#pragma GCC unroll 32for(int i=1;i<=n;i++)g[i].clear();#pragma GCC unroll 8for(int i=1;i<=n;i++){ok[i]=0;siz[i]=1;}#pragma GCC unroll 32fo(i, 2, n){qwq[i] = qr;Wadd(qwq[i], i);}#pragma GCC unroll 4fu(j, n, 2) siz[qwq[j]] += siz[j];int l = 1, r = n, ke = 0;while(l <= r){int mid = (l + r) >> 1;Wck(1, mid);if(f[1] == 0) r = mid - 1, ke = mid;else l = mid + 1;}Wck(1, ke - 1);Wsol(1);#pragma GCC unroll 32fo(i, 1, n){if(siz[i] - 1 > ke) putchar('1');else if(siz[i] - 1 < ke) putchar('0');else printf("%d", ok[i]);}puts("");}return Ratio;}
}
signed main(){return Wisadel::main();}
// Now there's only one thing I can do
// Fight until the end like I promised to