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。
4 3 3
1 5 1
1 5 1
3 5 2
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;
}
