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多校A层冲刺NOIP2024模拟赛11

时间：2024-10-22 17:00:13浏览次数：6  
标签：11 pp int 多校 Add mi1 mi2 NOIP2024 first

多校A层冲刺NOIP2024模拟赛11

\(T1\) A. 冒泡排序 \(100pts/100pts/100pts\)

  • 将循环 \(j\) 提到外面，本质上是对 \(a_{j},a_{j+k},a_{j+2k}, \dots ,a_{j+xk}\) 进行排序迭代的过程。

  • 按下标模 \(k\) 的余数分别排序即可。

    点击查看代码 
    int a[1000010];
vector<int>b[1000010];
int main()
{
  freopen("bubble.in","r",stdin);
  freopen("bubble.out","w",stdout);
  int n,k,i,j;
  scanf("%d%d",&n,&k);
  for(i=0;i<=n-1;i++)
  {
    scanf("%d",&a[i]);
    b[i%k].push_back(a[i]);
  }
  for(i=0;i<=k-1;i++)
  {
    sort(b[i].begin(),b[i].end());
    for(j=0;j<b[i].size();j++)
    {
      a[j*k+i]=b[i][j];
    }
  }
  for(i=0;i<=n-1;i++)
  {
    printf("%d ",a[i]);
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}

\(T2\) B. 染色 \(0pts/0pts/0pts\)

  • 难以保证最终得到的序列 \(\{ c' \}\) 本质不同，考虑正难则反，判断 \(\{ c' \}\) 能否被 \(\{ c \}\) 通过一系列操作得到。

  • 一个比较显然的结论是对于 \(\{ c' \}\) ，将同一个颜色段内的数都缩成一个数，设得到的序列为 \(\{ d \}\) ，那么 \(\{ c' \}\) 能被 \(\{ c \}\) 通过一系列操作得到当且仅当 \(\{ d \}\) 是 \(\{ c \}\) 的子序列。

    • 虽然赛时没想到。

  • 考虑统计 \(\{ c \}\) 满足相邻两项不能相同的本质不同子序列数量。设 \(F(i)\) 表示相邻两项不能相同的长度为 \(i\) 的本质不同子序列数量，由插板法有 \(\sum\limits_{i=1}^{n}\dbinom{n-1}{i-1}F(i)\) 即为所求。组合数判奇偶同 luogu P1869 愚蠢的组合数 ，难点在于如何求 \(F(i)\) 。

  • 设 \(f_{i,j,k}\) 表示前 \(i\) 个数中末尾元素为 \(j\) 且长度为 \(k\) 的相邻两项不能相同的的本质不同子序列数量，状态转移方程为 \(f_{i,j,k}=\begin{cases} [k=1]+\sum\limits_{h=1}^{m}f_{i-1,h,k-1} & c_{i}=j \\ f_{i-1,j,k} & c_{i} \ne j \end{cases}\) 。在使用 bitset 和前缀和优化后时间复杂度仍为 \(O(\frac{n^{2}m}{w})\) ，需要进一步优化。

  • 滚掉 \(i\) 这一维，有 \(f_{j,k}\) 表示表示末尾元素为 \(j\) 且长度为 \(k\) 的相邻两项不能相同的的本质不同子序列数量。

  • 在转移过程中发现有很多冗余转移（指 \(c_{i} \ne j\) 时），考虑对于每种元素单独考虑。具体地，对于 \(F(k)=\sum\limits_{j=1}^{m}f_{j,k}\) 记录一个总和 \(sum=\sum\limits_{k=1}^{n}F(k)\) 然后进行优化。

  • 而在二进制表示下， \(f_{j,k}=[k=1]+\sum\limits_{h=1}^{m}f_{h,k-1}\) 本质上是一个整体偏移的过程，使用 bitset 优化这一过程即可。

    点击查看代码 
    int c[200010];
bitset<100010>f[20010],tmp,sum;
int C(int n,int m)
{
	return (n&m)==m;
}
int main()
{
	freopen("color.in","r",stdin);
	freopen("color.out","w",stdout);
	int t,n,m,ans,i,j;
	scanf("%d",&t);
	for(j=1;j<=t;j++)
	{
		scanf("%d%d",&n,&m);	
		ans=0;
		for(i=1;i<=m;i++)
		{
			f[i].reset();
		}
		sum.reset();
		for(i=1;i<=n;i++)
		{
			scanf("%d",&c[i]);
		}
		for(i=1;i<=n;i++)
		{
			tmp=sum^f[c[i]];
			f[c[i]]=tmp<<1;
			f[c[i]][1]=1;
			sum=tmp^f[c[i]];
		}
		for(i=1;i<=n;i++)
		{
			ans^=C(n-1,i-1)*sum[i];
		}
		printf("%d",ans);
	}
	fclose(stdin);
	fclose(stdout);
	return 0;
}

\(T3\) C. 图 \(5pts/5pts/5pts\)

  • 部分分

    • 子任务 \(1\) ：暴力建边求最小生成树。

  • 正解

    • 抽象题目，直接挂官方题解了。

    点击查看不屑于封装的 std 代码 
    #include <cstdio>
#include <map>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <queue>
#include <stack>
#include <vector>
#include <random>
#include <cstring>
#include <ctime>
#include <cmath>
#include <assert.h>
#include <unordered_map>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define LL long long
#define pp pair<LL, LL>
#define mp make_pair
#define ull unsigned long long
namespace IO {
const int sz = 1 << 22;
char a[sz + 5], b[sz + 5], *p1 = a, *p2 = a, *t = b, p[105];
inline char gc() {
	//	return p1==p2?(p2=(p1=a)+fread(a,1,sz,stdin),p1==p2?EOF:*p1++):*p1++;
	return getchar();
}
template <class T>
void gi(T& x) {
	x = 0;
	int f = 1;
	char c = gc();
	if (c == '-')
		f = -1;
	for (; c < '0' || c > '9'; c = gc())
		if (c == '-')
			f = -1;
	for (; c >= '0' && c <= '9'; c = gc()) x = x * 10 + (c - '0');
	x = x * f;
}
inline void flush() { fwrite(b, 1, t - b, stdout), t = b; }
inline void pc(char x) {
	*t++ = x;
	if (t - b == sz)
		flush();
}
template <class T>
void pi(T x, char c = '\n') {
	if (x < 0)
		pc('-'), x = -x;
	if (x == 0)
		pc('0');
	int t = 0;
	for (; x; x /= 10) p[++t] = x % 10 + '0';
	for (; t; --t) pc(p[t]);
	pc(c);
}
struct F {
	~F() { flush(); }
} f;
}  // namespace IO
using IO::gi;
using IO::pc;
using IO::pi;
const int mod = 1e9 + 7;
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int dec(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline int mul(int x, int y) { return 1ll * x * y % mod; }
inline int qkpow(int a, int b) {
	if (b < 0)
		return 0;
	int ans = 1, base = a % mod;
	while (b) {
		if (b & 1)
			ans = 1ll * ans * base % mod;
		base = 1ll * base * base % mod;
		b >>= 1;
	}
	return ans;
}
int fac[1000005], inv[1000005], Invn[600005];
inline int binom(int n, int m) {
	if (n < m || m < 0)
		return 0;
	return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}
void init_C(int n) {
	fac[0] = 1;
	for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
	inv[0] = 1;
	inv[n] = qkpow(fac[n], mod - 2);
	for (int i = n - 1; i >= 1; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
	Invn[0] = Invn[1] = 1;
	for (int i = 1; i <= 200000; i++) Invn[i] = (LL)(mod - mod / i) * Invn[mod % i] % mod;
}
const LL INF = 1e18;
struct node3 {
	pp mi1, mi2;
	inline void init() { mi1 = mi2 = pp(INF, 0); }
} g1[100005], g2[100005], wg1[100005], wg2[100005];
inline void Add(node3& w1, pp w2) {
	if (!w2.second)
		return;
	if (w1.mi1.second == w2.second)
		w1.mi1.first = min(w1.mi1.first, w2.first);
	else if (w1.mi2.second == w2.second) {
		w1.mi2.first = min(w1.mi2.first, w2.first);
		if (w1.mi1.first > w1.mi2.first)
			swap(w1.mi1, w1.mi2);
	} else {
		if (w2.first < w1.mi1.first)
			w1.mi2 = w1.mi1, w1.mi1 = w2;
		else if (w2.first < w1.mi2.first)
			w1.mi2 = w2;
	}
}
int Log[100005];
struct ST_min {
	node3 f[100005][21];
	inline node3 query(int l, int r) {
		node3 res;
		res.init();
		if (l > r)
			return res;
		int k = Log[r - l + 1];
		res = f[r - (1 << k) + 1][k];
		Add(res, f[l][k].mi1);
		Add(res, f[l][k].mi2);
		return res;
	}
	inline void init(int N) {
		for (int j = 1; (1 << j) <= N; j++)
			for (int i = 1; i + (1 << j) - 1 <= N; i++) {
				f[i][j] = f[i + (1 << (j - 1))][j - 1];
				Add(f[i][j], f[i][j - 1].mi1);
				Add(f[i][j], f[i][j - 1].mi2);
			}
	}
} T1[2], T2[2];
int son1[100005], son2[100005], dep11[100005], dep22[100005], seg1[100005], seg2[100005], rev11[100005],
	rev22[100005];
int top1[100005], top2[100005];
int n, fa[200005], id1[100005], id2[100005], cnt1, cnt2, dfn1[100005], dfn2[100005];
int rev1[100005], rev2[100005], f2[100005][21], sz1[100005], sz2[100005], f1[100005][21];
int st1[100005][21], st2[100005][21];
LL jp1[100005][21], jp2[100005][21];
LL dep1[100005], dep2[100005];
pp mn[200005];
struct node {
	int to, w;
};
vector<node> G1[100005], G2[100005];
inline int findSet(int u) { return fa[u] == u ? u : fa[u] = findSet(fa[u]); }
inline int Min1(int u, int v) { return dfn1[u] < dfn1[v] ? u : v; }
inline int Min2(int u, int v) { return dfn2[u] < dfn2[v] ? u : v; }
inline int LCA1(int u, int v) {
	if (u == v)
		return u;
	if ((u = dfn1[u]) > (v = dfn1[v]))
		swap(u, v);
	int k = Log[v - u++];
	return Min1(f1[u][k], f1[v - (1 << k) + 1][k]);
}
inline int LCA2(int u, int v) {
	if (u == v)
		return u;
	if ((u = dfn2[u]) > (v = dfn2[v]))
		swap(u, v);
	int k = Log[v - u++];
	return Min2(f2[u][k], f2[v - (1 << k) + 1][k]);
}
inline LL getdis(int op, int u, int v) {
	if (op == 1)
		return dep1[u] + dep1[v] - 2 * dep1[LCA1(u, v)];
	else
		return dep2[u] + dep2[v] - 2 * dep2[LCA2(u, v)];
}
inline void dfs1(int u, int ff) {
	dep11[u] = dep11[ff] + 1;
	f1[dfn1[u] = ++cnt1][0] = ff;
	rev1[cnt1] = u;
	sz1[u] = 1;
	for (int i = 1; i <= 20; i++)
		st1[u][i] = st1[st1[u][i - 1]][i - 1], jp1[u][i] = jp1[u][i - 1] + jp1[st1[u][i - 1]][i - 1];
	for (auto to : G1[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		dep1[v] = dep1[u] + w;
		st1[v][0] = u, jp1[v][0] = w;
		dfs1(v, u);
		sz1[u] += sz1[v];
		if (sz1[v] > sz1[son1[u]])
			son1[u] = v;
	}
}
inline void dfs11(int u, int ff) {
	if (son1[u]) {
		seg1[son1[u]] = ++seg1[0];
		top1[son1[u]] = top1[u];
		rev11[seg1[0]] = son1[u];
		dfs11(son1[u], u);
	}
	for (auto to : G1[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		if (!top1[v]) {
			seg1[v] = ++seg1[0];
			top1[v] = v;
			rev11[seg1[0]] = v;
			dfs11(v, u);
		}
	}
}
inline void dfs2(int u, int ff) {
	dep22[u] = dep22[ff] + 1;
	f2[dfn2[u] = ++cnt2][0] = ff;
	rev2[cnt2] = u;
	sz2[u] = 1;
	for (int i = 1; i <= 20; i++)
		st2[u][i] = st2[st2[u][i - 1]][i - 1], jp2[u][i] = jp2[u][i - 1] + jp2[st2[u][i - 1]][i - 1];
	for (auto to : G2[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		dep2[v] = dep2[u] + w;
		st2[v][0] = u, jp2[v][0] = w;
		dfs2(v, u);
		sz2[u] += sz2[v];
		if (sz2[v] > sz2[son2[u]])
			son2[u] = v;
	}
}
inline void dfs22(int u, int ff) {
	if (son2[u]) {
		seg2[son2[u]] = ++seg2[0];
		top2[son2[u]] = top2[u];
		rev22[seg2[0]] = son2[u];
		dfs22(son2[u], u);
	}
	for (auto to : G2[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		if (!top2[v]) {
			seg2[v] = ++seg2[0];
			top2[v] = v;
			rev22[seg2[0]] = v;
			dfs22(v, u);
		}
	}
}
struct node2 {
	pp u, v;
} t[400005], po1[100005], po2[100005];
LL tag[400005];
#define ls(u) u << 1
#define rs(u) u << 1 | 1
inline node2 merge(int op, node2 A, node2 B) {
	node2 tmp;
	LL res = -1;
	LL d1 =
		(A.u.first == A.v.first ? A.u.second : A.u.second + A.v.second) + getdis(op, A.u.first, A.v.first);
	LL d2 =
		(B.u.first == B.v.first ? B.u.second : B.u.second + B.v.second) + getdis(op, B.u.first, B.v.first);
	LL d3 =
		(A.u.first == B.v.first ? A.u.second : A.u.second + B.v.second) + getdis(op, A.u.first, B.v.first);
	LL d4 =
		(B.u.first == A.v.first ? B.u.second : B.u.second + A.v.second) + getdis(op, B.u.first, A.v.first);
	LL d5 =
		(A.u.first == B.u.first ? A.u.second : A.u.second + B.u.second) + getdis(op, A.u.first, B.u.first);
	LL d6 =
		(A.v.first == B.v.first ? A.v.second : A.v.second + B.v.second) + getdis(op, A.v.first, B.v.first);
	res = max(d1, d2), res = max(res, max(d3, d4)), res = max(res, max(d5, d6));
	if (d1 == res)
		tmp = node2{ A.u, A.v };
	if (d2 == res)
		tmp = node2{ B.u, B.v };
	if (d3 == res)
		tmp = node2{ A.u, B.v };
	if (d4 == res)
		tmp = node2{ B.u, A.v };
	if (d5 == res)
		tmp = node2{ A.u, B.u };
	if (d6 == res)
		tmp = node2{ A.v, B.v };
	return tmp;
}
inline void push_down(int u) {
	if (tag[u]) {
		tag[ls(u)] += tag[u];
		tag[rs(u)] += tag[u];
		t[ls(u)].u.second += tag[u];
		t[ls(u)].v.second += tag[u];
		t[rs(u)].u.second += tag[u];
		t[rs(u)].v.second += tag[u];
		tag[u] = 0;
	}
}
inline void updata(int op, int p, int l, int r, int L, int R, int w) {
	if (L > R)
		return;
	if (L <= l && r <= R) {
		tag[p] += w;
		t[p].u.second += w;
		t[p].v.second += w;
		return;
	}
	push_down(p);
	int mid = (l + r) >> 1;
	if (L <= mid)
		updata(op, ls(p), l, mid, L, R, w);
	if (mid + 1 <= R)
		updata(op, rs(p), mid + 1, r, L, R, w);
	t[p] = merge(op, t[ls(p)], t[rs(p)]);
}
inline void build(int op, int p, int l, int r) {
	tag[p] = 0;
	if (l == r) {
		if (op == 2)
			t[p].u = t[p].v = pp(rev1[l], dep1[rev1[l]]);
		else
			t[p].u = t[p].v = pp(rev2[l], dep2[rev2[l]]);
		return;
	}
	int mid = (l + r) >> 1;
	build(op, ls(p), l, mid);
	build(op, rs(p), mid + 1, r);
	t[p] = merge(op, t[ls(p)], t[rs(p)]);
}
inline void redfs1(int u, int ff) {
	po1[u] = t[1];
	for (auto to : G1[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		updata(2, 1, 1, n, dfn1[v], dfn1[v] + sz1[v] - 1, -w);
		updata(2, 1, 1, n, 1, dfn1[v] - 1, w);
		updata(2, 1, 1, n, dfn1[v] + sz1[v], n, w);
		redfs1(v, u);
		updata(2, 1, 1, n, dfn1[v], dfn1[v] + sz1[v] - 1, w);
		updata(2, 1, 1, n, 1, dfn1[v] - 1, -w);
		updata(2, 1, 1, n, dfn1[v] + sz1[v], n, -w);
	}
}
inline void redfs2(int u, int ff) {
	po2[u] = t[1];
	for (auto to : G2[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		updata(1, 1, 1, n, dfn2[v], dfn2[v] + sz2[v] - 1, -w);
		updata(1, 1, 1, n, 1, dfn2[v] - 1, w);
		updata(1, 1, 1, n, dfn2[v] + sz2[v], n, w);
		redfs2(v, u);
		updata(1, 1, 1, n, dfn2[v], dfn2[v] + sz2[v] - 1, w);
		updata(1, 1, 1, n, 1, dfn2[v] - 1, -w);
		updata(1, 1, 1, n, dfn2[v] + sz2[v], n, -w);
	}
}
inline void dfss1(int u, int ff) {
	g1[u].init();
	wg1[u].init();
	Add(g1[u], pp(0, id1[u]));
	for (auto to : G1[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		dfss1(v, u);
		Add(g1[u], pp(g1[v].mi1.first + w, g1[v].mi1.second));
		Add(g1[u], pp(g1[v].mi2.first + w, g1[v].mi2.second));
	}
}
inline void redfss1(int u, int ff) {
	node3 pre;
	pre.init();
	Add(wg1[u], pp(0, id1[u]));
	for (auto to : G1[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		Add(wg1[v], pp(wg1[u].mi1.first + w, wg1[u].mi1.second));
		Add(wg1[v], pp(wg1[u].mi2.first + w, wg1[u].mi2.second));
		Add(wg1[v], pp(pre.mi1.first + w, pre.mi1.second));
		Add(wg1[v], pp(pre.mi2.first + w, pre.mi2.second));
		Add(pre, pp(g1[v].mi1.first + w, g1[v].mi1.second));
		Add(pre, pp(g1[v].mi2.first + w, g1[v].mi2.second));
	}
	reverse(G1[u].begin(), G1[u].end());
	pre.init();
	for (auto to : G1[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		Add(wg1[v], pp(pre.mi1.first + w, pre.mi1.second));
		Add(wg1[v], pp(pre.mi2.first + w, pre.mi2.second));
		Add(pre, pp(g1[v].mi1.first + w, g1[v].mi1.second));
		Add(pre, pp(g1[v].mi2.first + w, g1[v].mi2.second));
	}
	reverse(G1[u].begin(), G1[u].end());
	for (auto to : G1[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		redfss1(v, u);
	}
}
inline void dfss2(int u, int ff) {
	g2[u].init();
	wg2[u].init();
	Add(g2[u], pp(0, id2[u]));
	for (auto to : G2[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		dfss2(v, u);
		Add(g2[u], pp(g2[v].mi1.first + w, g2[v].mi1.second));
		Add(g2[u], pp(g2[v].mi2.first + w, g2[v].mi2.second));
	}
}
inline void redfss2(int u, int ff) {
	node3 pre;
	pre.init();
	Add(wg2[u], pp(0, id2[u]));
	for (auto to : G2[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		Add(wg2[v], pp(wg2[u].mi1.first + w, wg2[u].mi1.second));
		Add(wg2[v], pp(wg2[u].mi2.first + w, wg2[u].mi2.second));
		Add(wg2[v], pp(pre.mi1.first + w, pre.mi1.second));
		Add(wg2[v], pp(pre.mi2.first + w, pre.mi2.second));
		Add(pre, pp(g2[v].mi1.first + w, g2[v].mi1.second));
		Add(pre, pp(g2[v].mi2.first + w, g2[v].mi2.second));
	}
	reverse(G2[u].begin(), G2[u].end());
	pre.init();
	for (auto to : G2[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		Add(wg2[v], pp(pre.mi1.first + w, pre.mi1.second));
		Add(wg2[v], pp(pre.mi2.first + w, pre.mi2.second));
		Add(pre, pp(g2[v].mi1.first + w, g2[v].mi1.second));
		Add(pre, pp(g2[v].mi2.first + w, g2[v].mi2.second));
	}
	reverse(G2[u].begin(), G2[u].end());
	for (auto to : G2[u]) {
		int v = to.to, w = to.w;
		if (v == ff)
			continue;
		redfss2(v, u);
	}
}
inline node3 query22(int op, int u, int v, LL ex) {
	int fu = top2[u], fv = top2[v];
	node3 res;
	res.init();
	while (fu != fv) {
		if (dep22[fu] >= dep22[fv]) {
			node3 tmp = T2[op].query(seg2[fu], seg2[u]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
			u = st2[fu][0];
		} else {
			node3 tmp = T2[op].query(seg2[fv], seg2[v]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
			v = st2[fv][0];
		}
		fu = top2[u], fv = top2[v];
	}
	if (dep22[u] > dep22[v])
		swap(u, v);
	node3 tmp = T2[op].query(seg2[u], seg2[v]);
	Add(res, tmp.mi1), Add(res, tmp.mi2);
	return node3{ pp(res.mi1.first + ex, res.mi1.second), pp(res.mi2.first + ex, res.mi2.second) };
}
inline pp query2(int u, int v, LL w1, LL w2, int c) {  //��ɫ������ c
	int lca = LCA2(u, v);
	//	if(lca!=1)cerr<<lca<<" "<<"FUCK"<<endl;
	int to = u;
	LL sum = w1, tot = getdis(2, u, v) + w1 + w2;
	node3 res, tmp;
	res.init();
	LL ex = max(getdis(2, u, lca) + w1, getdis(2, v, lca) + w2);
	Add(res, pp(wg2[lca].mi1.first + ex, wg2[lca].mi1.second));
	Add(res, pp(wg2[lca].mi2.first + ex, wg2[lca].mi2.second));
	if (w1 >= tot - w1) {
		node3 tmp;
		tmp = query22(1, u, lca, w1 + dep2[u]);
		Add(res, tmp.mi1), Add(res, tmp.mi2);
	} else {
		for (int i = 20; i >= 0; i--) {
			if (sum + jp2[to][i] <= tot - sum - jp2[to][i]) {
				sum += jp2[to][i];
				to = st2[to][i];
			}
		}
		node3 tmp;
		if (dep22[to] <= dep22[lca]) {  //ȫ���� v ��
			tmp = query22(0, u, lca, w2 + dep2[v] - 2 * dep2[lca]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
		} else {
			tmp = query22(0, u, to, w2 + dep2[v] - 2 * dep2[lca]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
			tmp = query22(1, st2[to][0], lca, w1 + dep2[u]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
		}
	}
	to = v;
	sum = w2;
	if (w2 >= tot - w2) {
		tmp = query22(1, v, lca, w2 + dep2[v]);
		Add(res, tmp.mi1), Add(res, tmp.mi2);
	} else {
		for (int i = 20; i >= 0; i--) {
			if (sum + jp2[to][i] <= tot - sum - jp2[to][i]) {
				sum += jp2[to][i];
				to = st2[to][i];
			}
		}
		if (dep22[to] <= dep22[lca]) {  //ȫ���� u ��
			tmp = query22(0, v, lca, w1 + dep2[u] - 2 * dep2[lca]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
		} else {
			tmp = query22(0, v, to, w1 + dep2[u] - 2 * dep2[lca]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
			tmp = query22(1, st2[to][0], lca, w2 + dep2[v]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
		}
	}
	if (res.mi1.second == c)
		return res.mi2;
	return res.mi1;
}
inline node3 query11(int op, int u, int v, LL ex) {
	int fu = top1[u], fv = top1[v];
	node3 res;
	res.init();
	while (fu != fv) {
		if (dep11[fu] >= dep11[fv]) {
			node3 tmp = T1[op].query(seg1[fu], seg1[u]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
			u = st1[fu][0];
		} else {
			node3 tmp = T1[op].query(seg1[fv], seg1[v]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
			v = st1[fv][0];
		}
		fu = top1[u], fv = top1[v];
	}
	if (dep11[u] > dep11[v])
		swap(u, v);
	node3 tmp = T1[op].query(seg1[u], seg1[v]);
	Add(res, tmp.mi1), Add(res, tmp.mi2);
	return node3{ pp(res.mi1.first + ex, res.mi1.second), pp(res.mi2.first + ex, res.mi2.second) };
}
inline pp query1(int u, int v, LL w1, LL w2, int c) {  //��ɫ������ c
	int lca = LCA1(u, v);
	//	if(lca!=1)cerr<<lca<<" "<<"FUCK"<<endl;
	int to = u;
	LL sum = w1, tot = getdis(1, u, v) + w1 + w2;
	node3 res, tmp;
	res.init();
	LL ex = max(getdis(1, u, lca) + w1, getdis(1, v, lca) + w2);
	Add(res, pp(wg1[lca].mi1.first + ex, wg1[lca].mi1.second));
	Add(res, pp(wg1[lca].mi2.first + ex, wg1[lca].mi2.second));
	if (w1 >= tot - w1) {
		tmp = query11(1, u, lca, w1 + dep1[u]);
		Add(res, tmp.mi1), Add(res, tmp.mi2);
	} else {
		for (int i = 20; i >= 0; i--) {
			if (sum + jp1[to][i] <= tot - sum - jp1[to][i]) {
				sum += jp1[to][i];
				to = st1[to][i];
			}
		}
		node3 tmp;
		if (dep11[to] <= dep11[lca]) {  //ȫ���� v ��
			tmp = query11(0, u, lca, w2 + dep1[v] - 2 * dep1[lca]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
		} else {
			tmp = query11(0, u, to, w2 + dep1[v] - 2 * dep1[lca]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
			tmp = query11(1, st1[to][0], lca, w1 + dep1[u]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
		}
	}
	to = v;
	sum = w2;
	if (w2 >= tot - w2) {
		node3 tmp;
		tmp = query11(1, v, lca, w2 + dep1[v]);
		Add(res, tmp.mi1), Add(res, tmp.mi2);
	} else {
		for (int i = 20; i >= 0; i--) {
			if (sum + jp1[to][i] <= tot - sum - jp1[to][i]) {
				sum += jp1[to][i];
				to = st1[to][i];
			}
		}
		if (dep11[to] <= dep11[lca]) {  //ȫ���� u ��
			tmp = query11(0, v, lca, w1 + dep1[u] - 2 * dep1[lca]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
		} else {
			tmp = query11(0, v, to, w1 + dep1[u] - 2 * dep1[lca]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
			tmp = query11(1, st1[to][0], lca, w2 + dep1[v]);
			Add(res, tmp.mi1), Add(res, tmp.mi2);
		}
	}
	if (res.mi1.second == c)
		return res.mi2;
	return res.mi1;
}
inline void solve() {
	for (int i = 2; i <= 100000; i++) Log[i] = Log[i >> 1] + 1;
	gi(n);
	for (int i = 1; i < n; i++) {
		int u, v, w;
		gi(u), gi(v), gi(w);
		G1[u].push_back(node{ v, w });
		G1[v].push_back(node{ u, w });
	}
	for (int i = 1; i < n; i++) {
		int u, v, w;
		gi(u), gi(v), gi(w);
		G2[u].push_back(node{ v, w });
		G2[v].push_back(node{ u, w });
	}
	seg1[0] = seg1[1] = top1[1] = rev11[1] = 1;
	seg2[0] = seg2[1] = top2[1] = rev22[1] = 1;
	dfs1(1, 0), dfs2(1, 0), dfs11(1, 0), dfs22(1, 0);
	for (int j = 1; (1 << j) <= cnt1; j++)
		for (int i = 1; i + (1 << j) - 1 <= cnt1; i++)
			f1[i][j] = Min1(f1[i][j - 1], f1[i + (1 << (j - 1))][j - 1]);
	for (int j = 1; (1 << j) <= cnt2; j++)
		for (int i = 1; i + (1 << j) - 1 <= cnt2; i++)
			f2[i][j] = Min2(f2[i][j - 1], f2[i + (1 << (j - 1))][j - 1]);
	build(2, 1, 1, n);
	redfs1(1, 0);
	build(1, 1, 1, n);
	redfs2(1, 0);
	for (int i = 1; i <= 2 * n; i++) fa[i] = i;
	int cnt = 2 * n;
	LL ans = 0;
	while (cnt > 1) {
		for (int i = 1; i <= n; i++) id1[i] = findSet(i), mn[i] = pp(INF, 0);
		for (int i = 1; i <= n; i++) id2[i] = findSet(i + n), mn[i + n] = pp(INF, 0);
		dfss1(1, 0), dfss2(1, 0);
		redfss1(1, 0), redfss2(1, 0);
		for (int i = 1; i <= n; i++) {
			int u = rev11[i], v = rev22[i];
			T1[0].f[i][0].mi1 = pp(g1[u].mi1.first + dep1[u], g1[u].mi1.second);
			T1[0].f[i][0].mi2 = pp(g1[u].mi2.first + dep1[u], g1[u].mi2.second);
			T1[1].f[i][0].mi1 = pp(g1[u].mi1.first - dep1[u], g1[u].mi1.second);
			T1[1].f[i][0].mi2 = pp(g1[u].mi2.first - dep1[u], g1[u].mi2.second);

			T2[0].f[i][0].mi1 = pp(g2[v].mi1.first + dep2[v], g2[v].mi1.second);
			T2[0].f[i][0].mi2 = pp(g2[v].mi2.first + dep2[v], g2[v].mi2.second);
			T2[1].f[i][0].mi1 = pp(g2[v].mi1.first - dep2[v], g2[v].mi1.second);
			T2[1].f[i][0].mi2 = pp(g2[v].mi2.first - dep2[v], g2[v].mi2.second);
		}
		T1[0].init(n), T2[0].init(n), T1[1].init(n), T2[1].init(n);
		for (int i = 1; i <= n; i++) {
			int u = po1[i].u.first, v = po1[i].v.first;
			LL w1 = po1[i].u.second, w2 = po1[i].v.second;
			mn[id1[i]] = min(mn[id1[i]], query2(u, v, w1, w2, id1[i]));
		}
		for (int i = 1; i <= n; i++) {
			int u = po2[i].u.first, v = po2[i].v.first;
			LL w1 = po2[i].u.second, w2 = po2[i].v.second;
			mn[id2[i]] = min(mn[id2[i]], query1(u, v, w1, w2, id2[i]));
		}
		for (int i = 1; i <= 2 * n; i++) {
			if (mn[i].second) {
				int u = i, v = mn[i].second;
				u = findSet(u), v = findSet(v);
				if (u != v) {
					fa[v] = u;
					ans += mn[i].first;
					cnt--;
				}
			}
		}
		//	cerr<<cnt<<endl;
	}
	pi(ans);
}
/*
Ҫһ������
*/
signed main() {
	freopen("graph.in", "r", stdin);
	freopen("graph.out", "w", stdout);
	srand(time(0));
	solve();
	return 0;
}
/*
*/

\(T4\) D. 山峦 \(40pts/40pts/35pts\)

  • 部分分

    • 测试点 \(1,2,3,10,11,12\) ：爆搜。

      点击查看代码 
      const ll p=998244353;
ll h[15][15],c[15],ans[310];
void dfs(ll x,ll y,ll n,ll m,ll sum)
{
	if(x>n)
	{
		return;
	}
	if(x==n&&y==c[x])
	{
		if(y==1)
		{
			h[x][y]=c[x];
			if(sum+h[x][y]<=m)
			{
				ans[sum+h[x][y]]=(ans[sum+h[x][y]]+1)%p;	
			}
		}
		else
		{
			for(ll i=1;i<=min(h[x-1][y],h[x][y-1])&&sum+i<=m;i++)
			{
				h[x][y]=i;
				ans[sum+i]=(ans[sum+i]+1)%p;
			}
		}
	}
	else
	{
		if(y==1)
		{
			h[x][y]=c[x];
			if(sum+h[x][y]<=m)
			{
				if(y==c[x])
				{
					dfs(x+1,1,n,m,sum+h[x][y]);
				}
				else
				{
					dfs(x,y+1,n,m,sum+h[x][y]);
				}
			}
		}
		else
		{
			for(ll i=1;i<=min(h[x-1][y],h[x][y-1])&&sum+i<=m;i++)
			{
				h[x][y]=i;
				if(y==c[x])
				{
					dfs(x+1,1,n,m,sum+h[x][y]);
				}
				else
				{
					dfs(x,y+1,n,m,sum+h[x][y]);
				}
			}
		}
	}
}
int main()
{
	freopen("mountain.in","r",stdin);
	freopen("mountain.out","w",stdout);
	ll n,m,i;
	scanf("%lld%lld",&n,&m);
	for(i=1;i<=n;i++)
	{
		scanf("%lld",&c[i]);
	}
	memset(h[0],0x3f,sizeof(h[0]));
	dfs(1,1,n,m,0);
	for(i=1;i<=m;i++)
	{
		printf("%lld ",ans[i]);
	}
	fclose(stdin);
	fclose(stdout);
	return 0;
}

  • 正解

    • 不会高维前缀和，先咕了。

    点击查看代码 
    #include <cstdio>
#include <map>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <queue>
#include <stack>
#include <cstring>
#include <ctime>
#include <cmath>
#include <assert.h>
using namespace std;
#define LL long long
#define pp pair<int, int>
#define mp make_pair
#define ull unsigned long long
namespace IO {
const int sz = 1 << 22;
char a[sz + 5], b[sz + 5], *p1 = a, *p2 = a, *t = b, p[105];
inline char gc() {
	//	return p1==p2?(p2=(p1=a)+fread(a,1,sz,stdin),p1==p2?EOF:*p1++):*p1++;
	return getchar();
}
template <class T>
void gi(T& x) {
	x = 0;
	int f = 1;
	char c = gc();
	if (c == '-')
		f = -1;
	for (; c < '0' || c > '9'; c = gc())
		if (c == '-')
			f = -1;
	for (; c >= '0' && c <= '9'; c = gc()) x = x * 10 + (c - '0');
	x = x * f;
}
inline void flush() { fwrite(b, 1, t - b, stdout), t = b; }
inline void pc(char x) {
	*t++ = x;
	if (t - b == sz)
		flush();
}
template <class T>
void pi(T x, char c = '\n') {
	if (x < 0)
		pc('-'), x = -x;
	if (x == 0)
		pc('0');
	int t = 0;
	for (; x; x /= 10) p[++t] = x % 10 + '0';
	for (; t; --t) pc(p[t]);
	pc(c);
}
struct F {
	~F() { flush(); }
} f;
}  // namespace IO
using IO::gi;
using IO::pc;
using IO::pi;
const int mod = 998244353;
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int dec(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline int qkpow(int a, LL b) {
	if (b < 0)
		return 0;
	int ans = 1, base = a % mod;
	while (b) {
		if (b & 1)
			ans = 1ll * ans * base % mod;
		base = 1ll * base * base % mod;
		b >>= 1;
	}
	return ans;
}
int fac[1000005], inv[1000005], Invn[600005];
inline int C(int n, int m) {
	if (n < m || m < 0)
		return 0;
	return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}
void init_C(int n) {
	fac[0] = 1;
	for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
	inv[0] = 1;
	inv[n] = qkpow(fac[n], mod - 2);
	for (int i = n - 1; i >= 1; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
	Invn[0] = Invn[1] = 1;
	for (int i = 1; i <= 200000; i++) Invn[i] = (LL)(mod - mod / i) * Invn[mod % i] % mod;
}
int n, h, l[15], b[25], Len[15], res;
LL pw[15];
int dp[11][60005][305], pre[60005][305];
map<LL, int> Id[15];
struct node {
	int id;
	int w, b[15];
} a[11][60005];
vector<pp> G[15][15][15];
inline void dfs(int lim, int y, int tot, int lst) {
	if (tot > n)
		return;
	if (y == 0) {
		Len[lim]++;
		a[lim][Len[lim]].w = tot;
		LL S = 0;
		for (int i = 2; i <= lim; i++) S += 1ll * pw[i - 2] * (b[i] - 1);
		a[lim][Len[lim]].id = S;
		for (int i = 2; i <= lim; i++) a[lim][Len[lim]].b[i] = b[i] - 1;
		Id[lim][S] = Len[lim];
		return;
	}
	if (y == 1)
		b[y] = lim, dfs(lim, y - 1, tot, 1);
	else {
		for (int i = lst; i <= lim; i++) {
			b[y] = i;
			dfs(lim, y - 1, tot + i, i);
		}
	}
}
signed main() {
	freopen("mountain.in", "r", stdin);
	freopen("mountain.out", "w", stdout);
	pw[0] = 1;
	for (int i = 1; i <= 10; i++) pw[i] = 1ll * pw[i - 1] * 9;
	gi(h), gi(n);
	int ex = 0;
	for (int i = 1; i <= h; i++) gi(l[i]), ex += l[i];
	if (ex > n) {
		for (int i = 1; i <= n; i++) pi(0, ' ');
		return 0;
	}
	n -= ex;
	for (int i = 1; i <= 10; i++) {
		dfs(i, i, 0, 1);
		for (int j = 2; j <= i; j++) {
			for (int k = 1; k <= Len[i]; k++) {
				int x = a[i][k].b[j];
				LL nS = 0;
				if (x == i - 1)
					continue;
				for (int p = 2; p <= i; p++) {
					if (p < j)
						nS += 1ll * pw[p - 2] * max(a[i][k].b[p], x + 1);
					else if (p == j)
						nS += 1ll * pw[p - 2] * (x + 1);
					else
						nS += 1ll * pw[p - 2] * a[i][k].b[p];
				}
				if (Id[i].find(nS) != Id[i].end())
					G[i][j][x].push_back(pp(k, Id[i][nS]));
			}
		}
	}
	for (int i = 1; i <= h; i++) {
		if (i == 1) {
			for (int j = 1; j <= Len[l[i]]; j++) dp[i][j][a[l[i]][j].w]++;
		} else {
			for (int j = 1; j <= Len[l[i]]; j++)
				for (int k = 0; k <= n; k++) pre[j][k] = 0;
			for (int j = 1; j <= Len[l[i - 1]]; j++) {
				LL S = 0;
				for (int k = 2; k <= l[i]; k++) S += 1ll * pw[k - 2] * min(a[l[i - 1]][j].b[k], l[i] - 1);
				S = Id[l[i]][S];
				for (int k = 0; k <= n; k++) pre[S][k] = add(pre[S][k], dp[i - 1][j][k]);
			}
			for (int j = 2; j <= l[i]; j++) {
				for (int k = l[i] - 1; k >= 0; k--) {
					for (int p = 0; p < G[l[i]][j][k].size(); p++) {
						int id1 = G[l[i]][j][k][p].first;
						int id2 = G[l[i]][j][k][p].second;
						for (int p2 = 0; p2 <= n; p2++) pre[id1][p2] = add(pre[id1][p2], pre[id2][p2]);
					}
				}
			}
			for (int j = 1; j <= Len[l[i]]; j++) {
				int w = a[l[i]][j].w;
				for (int k = 0; k <= n - w; k++) {
					dp[i][j][k + w] = add(dp[i][j][k + w], pre[j][k]);
				}
			}
		}
	}
	for (int i = 1; i < ex; i++) pi(0, ' ');
	for (int p = 0; p <= n; p++) {
		res = 0;
		for (int i = 1; i <= Len[l[h]]; i++) res = add(res, dp[h][i][p]);
		pi(res, ' ');
	}
	return 0;
}
/*
2 2
2 2
*/

总结

  • \(T2\) 想到正难则反，但没发现重要结论，不会设状态。
  • \(T3\) 测 \(n \le 2000\) 的数据时，将树剖求 \(\operatorname{LCA}\) 换成了 \(DFS\) 序求 \(\operatorname{LCA}\) ，但优化不明显（下发的样例仅快了 \(2s\)） 。

标签：11,pp,int,多校,Add,mi1,mi2,NOIP2024,first
From： https://www.cnblogs.com/The-Shadow-Dragon/p/18493299

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