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;

}

巴特西

HNCPC2019部分题解

C. Distinct Substrings

D. Modulo Nine

Code

G. 字典序

Code

H. 有向图

Code

J. Parity of Tuples (Easy)

Code

K. 双向链表练习题

Code

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