HNCPC2019部分题解

签到题就不写了。

C. Distinct Substrings

先对原串建出SAM，map存边。

由于这题相当于添加一个字符再删除这个字符，添加下一个字符，所以每次都暴力跳后缀链接是复杂度是错的。

从 $last$ 向上跳的时候，遍历出边，把每个拥有数字 $c$ 的出边的第一个点编号记下来，由于后缀自动机的边数是线性的，这样复杂度就是对的 $O(n\log n)$ 虽然 $5e6$ 但数据水还是跑过去了。

#include <iostream>
#include <map>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <fstream>
typedef long long LL;
typedef unsigned long long uLL;
#define Int __int128
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define MP(x, y) std::make_pair(x, y)
#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define GO cerr << "GO" << endl;
using namespace std;
inline void proc_status()
{
	ifstream t("/proc/self/status");
	cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
template<class T> inline T read()
{
	register int x = 0; register int f = 1; register char c;
	while (!isdigit(c = getchar())) if (c == '-') f = -1;
	while (x = (x << 1) + (x << 3) + (c xor 48), isdigit(c = getchar()));
	return x * f;
}
template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
const int maxN = (int) 1e6;
const int mod = (int) 1e9 + 7;
int pw[maxN + 2];
int ncnt, last, n, m;
struct Status
{
	int len, link;
	map<int, int> ch;
} st[2 * maxN];
void init() { ncnt = 0, last = 0, st[0].len = 0, st[0].link = -1; }
void insert(int c)
{
	int cur = ++ncnt;
	int p = last;
	st[cur].len = st[last].len + 1;
	while (p != -1 and !st[p].ch.count(c))
	{
		st[p].ch[c] = cur;
		p = st[p].link;
	}
	if (p == -1)
		st[cur].link = 0;
	else
	{
		int q = st[p].ch[c];
		if (st[q].len == st[p].len + 1)
			st[cur].link = q;
		else
		{
			int clone = ++ncnt;
			st[clone] = st[q];
			st[clone].len = st[p].len + 1;
			while (p != -1 and st[p].ch[c] == q)
			{
				st[p].ch[c] = clone;
				p = st[p].link;
			}
			st[cur].link = st[q].link = clone;
		}
	}
	last = cur;
}
void Clear()
{
	for (register int i = 0; i <= ncnt; ++i)
		st[i].ch.clear(), st[i].len = st[i].link = 0;
	ncnt = last = 0;
}
void Init()
{
	pw[0] = 1;
	for (register int i = 1; i <= m; ++i) pw[i] = 3ll * pw[i - 1] % mod;
}
void Solve()
{
	int ans = 0;
	static int place[maxN + 2];
	fill(place + 1, place + 1 + m, -1);
	int L = st[last].len + 1, p = last;
	while (p != -1)
	{
		for (map<int, int>::iterator it = st[p].ch.begin(); it != st[p].ch.end(); ++it)
		{
			if (place[it->first] < 0)
				place[it->first] = p;
		}
		p = st[p].link;
	}
	for (register int i = 1; i <= m; ++i)
	{
		if (place[i] == -1)
			ans ^= (LL) L * pw[i] % mod;
		else
			ans ^= (LL) (L - st[place[i]].len - 1) * pw[i] % mod;
	}
	cout << ans << endl;
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("C.in", "r", stdin);
	freopen("C.out", "w", stdout);
#endif
	while (scanf("%d%d", &n, &m) != EOF)
	{
		init();
		Init();
		for (register int i = 1, x; i <= n; ++i)
			x = read<int>(), insert(x);
		Solve();
		Clear();
	}
	return 0;
}

D. Modulo Nine

设 $dp[i][j][k]$ 表示填了前 $i$ 个位置，最近的一个含3这个因子的数在 $k$，次近的在 $j$的方案数。

那么对每个点求出一个最近的需要满足限制的左端点（没有就默认为0），如果状态 $dp[i][j][k]$ 满足 j和k都在i最近的左端点内，那么这个状态就是合法的，接下考虑i+1位填 $\{0, 9\}$ 或 $\{3, 6\}$ 或 $\{1, 2, 4, 5, 7, 8\}$，刷表转移即可。

Code

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
void chkmax(int &x, int y) { x < y ? x = y : 0; }
const int maxN = 50;
const int mod = (int) 1e9 + 7;
int n, m;
int L[maxN + 2];
int dp[maxN + 2][maxN + 2][maxN + 2];
void pls(int &x, int y)
{
	x += y;
	if (x >= mod) x -= mod;
	if (x < 0) x += mod;
}
int main()
{
	freopen("D.in", "r", stdin);
	freopen("D.out", "w", stdout);
	while (scanf("%d%d", &n, &m) != EOF)
	{
		memset(L, 0, sizeof L);
		for (int i = 1; i <= m; ++i)
		{
			int l, r;
			scanf("%d%d", &l, &r);
			chkmax(L[r], l);
		}
		memset(dp, 0, sizeof dp);
		dp[0][0][0] = 1;
		for (int i = 0; i < n; ++i)
			for (int j = 0; j <= i; ++j)
				for (int k = j; k <= i; ++k)
				{
					if (!dp[i][j][k]) continue;
					if (L[i] <= j and L[i] <= k)
					{
						int curr = dp[i][j][k];
						// {1, 2, 4, 5, 7, 8}
						pls(dp[i + 1][j][k], 6ll * curr % mod);
						// {3, 6}
						pls(dp[i + 1][k][i + 1], 2ll * curr % mod);
						// {0, 9}
						pls(dp[i + 1][i + 1][i + 1], 2ll * curr % mod);
					}
				}
		int ans = 0;
		for (int i = L[n]; i <= n; ++i)
			for (int j = i; j <= n; ++j)
				pls(ans, dp[n][i][j]);
		printf("%d\n", ans);
	}
}

G. 字典序

参考了zsy的题解：

不难发现只要考虑相邻两行的限制，如果满足任意相邻两行满足，那么就是合法的，即 $\forall i\in[1, n - 1],j\in[2,m],\exists k < j$ 满足 $a[i][j] > a[i + 1][j]$ 且 $a[i][k] < a[i + 1][k]$。

这个限制可以抽象成 $1\le i\le n - 1$，$\{1,2\cdots m\}$ 的两个子集 $A[i]$ 和 $B[i]$，需要满足最终的排列中 $B[i]$ 中的每一个元素都在至少一个 $A[i]$ 中元素的后面。

那么每个点（列）会有些限制，这些限制的个数就是被多少个B集合包含，因为B集合是被A集合限制的，当限制个数为0时，我们就可以把它加入到排列中了。

比如上面这个：$A[1] = \{2\},A[2] = \{\}, A[3]=\{1,3\}$, $B[1] = \{1, 3\}, B[2] = \{\}, B[3] = \{\}$

我们从前往后构造排列，若构造了 $A[x]$ 中的数，那么对应的 $B[x]$ 所包含的列的限制就会-1，所以每次找到最小的限制数为0的列，作为当前构造排列的最后一个，然后把它所在 $A[x]$ 集合对应的 $B[x]$ 集合中的列的限制-1，再将 $A[x]$ 和 $B[x]$ 删掉。

Code

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
const int maxN = 2000;
int n, m, a[maxN + 2][maxN + 2];
int limits[maxN + 2];
vector<int> B_to_Lie[maxN + 2];
vector<int> Lie_to_A[maxN + 2];
int main()
{
	freopen("G.in", "r", stdin);
	freopen("G.out", "w", stdout);
	while (scanf("%d%d", &n, &m) != EOF)
	{
		memset(limits, 0, sizeof limits);
		for (int i = 1; i < n; ++i) B_to_Lie[i].clear();
		for (int i = 1; i <= m; ++i) Lie_to_A[i].clear();
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= m; ++j)
				scanf("%d", &a[i][j]);
		for (int i = 1; i < n; ++i)
			for (int j = 1; j <= m; ++j)
				if (a[i][j] < a[i + 1][j])
					Lie_to_A[j].push_back(i);
				else if (a[i][j] > a[i + 1][j])
				{
					limits[j]++;
					B_to_Lie[i].push_back(j);
				}
		int tot = 0;
		static int ans[maxN + 2];
		for (int t = 1; t <= m; ++t)
		{
			int p = 0;
			for (int i = 1; i <= m; ++i)
				if (!limits[i])
				{
					p = i;
					limits[i] = 0x3f3f3f3f;
					break;
				}
			if (!p)
				break;
			ans[++tot] = p;
			for (int i = 0; i < (int) Lie_to_A[p].size(); ++i)
			{
				int A = Lie_to_A[p][i];
				for (int j = 0; j < (int) B_to_Lie[A].size(); ++j)
				{
					int Lie = B_to_Lie[A][j];
					limits[Lie]--;
				}
				B_to_Lie[A].clear();
			}
			Lie_to_A[p].clear();
		}
		if (tot == m)
			for (int i = 1; i <= m; ++i)
				if (i != m) printf("%d ", ans[i]);
				else printf("%d\n", ans[i]);
		else
			puts("-1");
	}
	return 0;
}

H. 有向图

设 $E[u]$ 为走无限次后期望经过 $u$ 多少次，由于经过n+1...n+m的次数为1，所以此时n+1...n+m点期望就是概率。

列出方程高消即可。

Code

#include <iostream>
#include <cstring>
#include <cstdio>
#define LL long long
using namespace std;
const int maxN = 500;
const int mod = (int) 1e9 + 7;
LL qpow(LL a, LL b)
{
	LL ans = 1;
	while (b)
	{
		if (b & 1)
			ans = ans * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return ans;
}
int n, m;
int a[maxN + 2][maxN + 2], P[maxN + 2][maxN + 2];
int main()
{
	freopen("H.in", "r", stdin);
	freopen("H.out", "w", stdout);
	while (scanf("%d%d", &n, &m) != EOF)
	{
		memset(a, 0, sizeof a);
		memset(P, 0, sizeof P);
		int inv = qpow(10000, mod - 2);
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n + m; ++j)
			{
				scanf("%d", &P[i][j]);
				P[i][j] = 1ll * P[i][j] * inv % mod;
			}
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= n; ++j)
			{
				a[i][j] = P[j][i];
				if (i == j) a[i][i]--;
				(a[i][j] += mod) %= mod;
			}
			if (i == 1) a[1][n + 1] = mod - 1;
			else a[i][n + 1] = 0;
		}
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= n; ++j)
				if (i != j)
				{
					int mul = 1ll * a[j][i] * qpow(a[i][i], mod - 2) % mod;
					for (int k = 1; k <= n + 1; ++k)
					{
						a[j][k] -= 1ll * a[i][k] * mul % mod;
						a[j][k] %= mod;
						(a[j][k] += mod) %= mod;
					}
				}
		}
		for (int i = 1; i <= n; ++i)
			a[i][n + 1] = (1ll * a[i][n + 1] * qpow(a[i][i], mod - 2) % mod + mod) % mod;
	   	for (int i = 1 + n; i <= n + m; ++i)
		{
			int ans = 0;
			for (int j = 1; j <= n; ++j)
				(ans += 1ll * a[j][n + 1] * P[j][i] % mod) %= mod;
			printf("%d", (ans + mod) % mod);
			if (i != n + m)
				putchar(' ');
		}
		putchar('\n');
	}
}

J. Parity of Tuples (Easy)

首先对每一行单独考虑贡献再加起来。

接下来都是对一行考虑：

由于要求and x后二进制下都是奇数个1，所以设 $f[i][S]$ 为考虑了 $x$ 的前i位，这一行每个数二进制下1个数的奇偶性二进制状态为S对答案的贡献。

对于 $3^x$ 将x拆成二进制相加，转移时乘上即可。

Code

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxN = (int) 1e4, maxM = 30, maxK = 30;
const int mod = (int) 1e9 + 7;
int n, m, k;
int b[maxK + 2], pw[maxK + 2];
int a[maxN + 2][maxM + 2];
int f[maxK + 2][1 << maxM];
int main()
{
	freopen("J.in", "r", stdin);
	freopen("J.out", "w", stdout);
	while (scanf("%d%d%d", &n, &m, &k) != EOF)
	{
		for (int i = 1; i <= n; ++i)
			for (int j = 0; j < m; ++j)
				scanf("%d", &a[i][j]);
		int ans = 0;
		for (int t = 1; t <= n; ++t)
		{
			for (int i = 0; i <= k; ++i)
			{
				b[i] = 0;
				for (int j = 0; j < m; ++j)
					b[i] |= (a[t][j] >> i & 1) << j;
			}
			pw[0] = 3;
			for (int i = 1; i <= k; ++i)
				pw[i] = 1ll * pw[i - 1] * pw[i - 1] % mod;
			memset(f, 0, sizeof f);
			f[0][0] = 1;
			for (int i = 0; i < k; ++i)
				for (int S = 0; S < (1 << m); ++S)
					if (f[i][S])
					{
						(f[i + 1][S ^ b[i]] += 1ll * f[i][S] * pw[i]) %= mod;
						(f[i + 1][S] += f[i][S]) %= mod;
					}
			(ans += f[k][(1 << m) - 1]) %= mod;
		}
		printf("%d\n", (ans + mod) % mod);
	}
}

K. 双向链表练习题

deque 启发式合并。

Code

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <fstream>
typedef long long LL;
typedef unsigned long long uLL;
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define MP(x, y) std::make_pair(x, y)
#define DE(x) cerr << x << endl;
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define GO cerr << "GO" << endl;
#define rep(i, a, b) for (register int (i) = (a); (i) <= (b); ++(i))
using namespace std;
inline void proc_status()
{
	ifstream t("/proc/self/status");
	cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
inline int read()
{
	register int x = 0; register int f = 1; register char c;
	while (!isdigit(c = getchar())) if (c == '-') f = -1;
	while (x = (x << 1) + (x << 3) + (c xor 48), isdigit(c = getchar()));
	return x * f;
}
template<class T> inline void write(T x)
{
	static char stk[30]; static int top = 0;
	if (x < 0) { x = -x, putchar('-'); }
	while (stk[++top] = x % 10 xor 48, x /= 10, x);
	while (putchar(stk[top--]), top);
}
template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
const int maxN = 2e5;
deque<int> q[maxN + 2];
int n, m;
bool rev[maxN + 2];
int size(int x) { return q[x].size(); }
int front(int x)
{
	if (rev[x]) return q[x].back();
	else return q[x].front();
}
int back(int x)
{
	if (rev[x]) return q[x].front();
	else return q[x].back();
}
void push_front(int x, int y)
{
	if (rev[x]) q[x].push_back(y);
	else q[x].push_front(y);
}
void push_back(int x, int y)
{
	if (rev[x]) q[x].push_front(y);
	else q[x].push_back(y);
}
void pop_back(int x)
{
	if (rev[x]) q[x].pop_front();
	else q[x].pop_back();
}
void pop_front(int x)
{
	if (rev[x]) q[x].pop_back();
	else q[x].pop_front();
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("K.in", "r", stdin);
	freopen("K.out", "w", stdout);
#endif
	while (scanf("%d%d", &n, &m) != EOF)
	{
		for (int i = 1; i <= n; ++i)
			q[i].clear(), q[i].push_front(i), rev[i] = 0;
		for (int i = 1; i <= m; ++i)
		{
			int a, b;
			scanf("%d%d", &a, &b);
			if (q[a].size() > q[b].size())
			{
				while (size(b))
				{
					push_back(a, front(b));
					pop_front(b);
				}
				rev[a] ^= 1;
			} else
			{
				while (size(a))
				{
					push_front(b, back(a));
					pop_back(a);
				}
				rev[a] = rev[b] ^ 1;
				swap(q[a], q[b]);
			}
		}
		printf("%d", q[1].size());
		if (rev[1]) { while (q[1].size()) printf(" %d", q[1].back()), q[1].pop_back(); }
		else { while (q[1].size()) printf(" %d", q[1].front()), q[1].pop_front(); }
		puts("");
	}
	return 0;
}