problem1 link
假设第$i$种出现的次数为$n_{i}$，总个数为$m$，那么排列数为$T=\frac{m!}{\prod_{i=1}^{26}(n_{i}!)}$
然后计算回文的个数，只需要考虑前一半，得到个数为$R$，那么答案为$\frac{R}{T}$.
为了防止数字太大导致越界，可以分解为质因子的表示方法。
problem2 link
假设终点所在的位置为$(tx,ty)$，那么所有底面是$1x1$的格子$(x,y)$一定满足$(x-tx)mod(3)=0,(y-ty)mod(3)=0$
把每个这样的点拆成两个点然后建立最小割的图。
源点与所有的$b$相连，终点与汇点相连，流量为无穷，割边不会在这里产生。
如果不是洞，那么这个格子拆成的两个点流量为1,表示将这个格子设为洞。
每个格子向周围连边，流量为将中间的两个格子设为洞的代价。
最后最小割就是答案。
problem3 link
首先考虑集合之间的关系。设$f[i][j]$表示前$i$个人分成$j$个集合的方案数。初始化$f[1][1]=n$。那么有：
(1)$f[i+1][j+1]=f[i][j]*j$表示新加一个集合，可以在任意两个集合之间
(2)$f[i+1][j]=f[i][j]*j*2$表示新加的元素与之前的某一个集合在一起，可以放在那个集合的前后，所以有$j*2$种方法
(３)$f[i+1][j-1]=f[i][j]*j$表示合并两个集合，可以在任意两个集合之间插入从而进行合并
最后就是对于$f[x][y]$来说，有多少种方式可以在$n$个位置上放置$x$个使得有$y$个集合并且任意两个集合不相邻。令$m=n-(x-y)$，那么相当于在$m$个位置中放置$y$个，使得任意两个不相邻。由于$f[1][1]=n$那么这$y$个集合的排列已经计算了，所以现在可以假设这$y$个元素的第一个放在$ｍ$个位置的第一个位置，那么第二个位置也不能放置了。所以还剩$m-2$个位置,$y-1$个元素。由于每放置一个元素其后面的位置就不能放置了，所以可以把剩下$y-1$个元素的位置与其后面相邻的位置绑定成一个位置，这样的话，就是$m-2-(y-1)$个位置，$y-1$个元素，即$C_{m-2-(y-1)}^{y-1}=C_{n-(x-y)-2-(y-1)}^{y-1}=C_{n-x-1}^{y-1}$
code for problem1
#include <cmath>
#include <string>
#include <vector> class PalindromePermutations {
public:
double palindromeProbability(const std::string &word) {
std::vector<int> h(26, 0);
for (auto e : word) {
++h[e - 'a'];
}
int old_idx = -1;
for (int i = 0; i < 26; ++i) {
if (h[i] % 2 == 1) {
if (old_idx != -1) {
return 0.0;
}
old_idx = i;
}
}
auto total = Compute(h);
if (old_idx != -1) {
--h[old_idx];
}
for (auto &e : h) {
e /= 2;
}
auto target = Compute(h);
double result = 1.0;
for (int i = 2; i < 50; ++i) {
result *= std::pow(i, target[i] - total[i]);
}
return result;
} private:
std::vector<int> Compute(const std::vector<int> &h) {
std::vector<int> result(50, 0);
auto Add = [&](int x, int sgn) {
for (int i = 2; i <= x; ++i) {
int k = i;
for (int j = 2; j * j <= k; ++j) {
while (k % j == 0) {
result[j] += sgn;
k /= j;
}
}
if (k != 1) {
result[k] += sgn;
}
}
};
int n = 0;
for (auto e : h) {
Add(e, -1);
n += e;
}
Add(n, 1);
return result;
}
};
code for problem2
#include <limits>
#include <unordered_map>
#include <vector> template <typename FlowType>
class MaxFlowSolver {
static constexpr FlowType kMaxFlow = std::numeric_limits<FlowType>::max();
static constexpr FlowType kZeroFlow = static_cast<FlowType>(0);
struct node {
int v;
int next;
FlowType cap;
}; public:
int VertexNumber() const { return used_index_; } FlowType MaxFlow(int source, int sink) {
source = GetIndex(source);
sink = GetIndex(sink); int n = VertexNumber();
std::vector<int> pre(n);
std::vector<int> cur(n);
std::vector<int> num(n);
std::vector<int> h(n);
for (int i = 0; i < n; ++i) {
cur[i] = head_[i];
num[i] = 0;
h[i] = 0;
}
int u = source;
FlowType result = 0;
while (h[u] < n) {
if (u == sink) {
FlowType min_cap = kMaxFlow;
int v = -1;
for (int i = source; i != sink; i = edges_[cur[i]].v) {
int k = cur[i];
if (edges_[k].cap < min_cap) {
min_cap = edges_[k].cap;
v = i;
}
}
result += min_cap;
u = v;
for (int i = source; i != sink; i = edges_[cur[i]].v) {
int k = cur[i];
edges_[k].cap -= min_cap;
edges_[k ^ 1].cap += min_cap;
}
}
int index = -1;
for (int i = cur[u]; i != -1; i = edges_[i].next) {
if (edges_[i].cap > 0 && h[u] == h[edges_[i].v] + 1) {
index = i;
break;
}
}
if (index != -1) {
cur[u] = index;
pre[edges_[index].v] = u;
u = edges_[index].v;
} else {
if (--num[h[u]] == 0) {
break;
}
int k = n;
cur[u] = head_[u];
for (int i = head_[u]; i != -1; i = edges_[i].next) {
if (edges_[i].cap > 0 && h[edges_[i].v] < k) {
k = h[edges_[i].v];
}
}
if (k + 1 < n) {
num[k + 1] += 1;
}
h[u] = k + 1;
if (u != source) {
u = pre[u];
}
}
}
return result;
} MaxFlowSolver() = default; void Clear() {
edges_.clear();
head_.clear();
vertex_indexer_.clear();
used_index_ = 0;
} void InsertEdge(int from, int to, FlowType cap) {
from = GetIndex(from);
to = GetIndex(to);
AddEdge(from, to, cap);
AddEdge(to, from, kZeroFlow);
} private:
int GetIndex(int idx) {
auto iter = vertex_indexer_.find(idx);
if (iter != vertex_indexer_.end()) {
return iter->second;
}
int map_idx = used_index_++;
head_.push_back(-1);
return vertex_indexer_[idx] = map_idx;
} void AddEdge(int from, int to, FlowType cap) {
node p;
p.v = to;
p.cap = cap;
p.next = head_[from];
head_[from] = static_cast<int>(edges_.size());
edges_.emplace_back(p);
} std::vector<node> edges_;
std::vector<int> head_; std::unordered_map<int, int> vertex_indexer_;
int used_index_ = 0;
}; class BlockTheBlockPuzzle {
static constexpr int kInfinite = 1000000; public:
int minimumHoles(const std::vector<std::string> &S) {
MaxFlowSolver<int> solver;
int n = static_cast<int>(S.size());
int source = -1;
int sink = -2;
int tx = -1, ty = -1;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (S[i][j] == '$') {
tx = i;
ty = j;
}
}
} auto P0 = [&](int i, int j) { return i * n + j; };
auto P1 = [&](int i, int j) { return i * n + j + n * n; }; for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i % 3 == tx % 3 && j % 3 == ty % 3) {
if (S[i][j] == '$') {
solver.InsertEdge(P1(i, j), sink, kInfinite);
}
if (S[i][j] == 'b') {
solver.InsertEdge(source, P0(i, j), kInfinite);
}
if (S[i][j] != 'H') {
solver.InsertEdge(P0(i, j), P1(i, j),
S[i][j] == '.' ? 1 : kInfinite);
}
if (i + 3 < n) {
auto cost = GetCost(S, i + 1, j, i + 2, j);
solver.InsertEdge(P1(i, j), P0(i + 3, j), cost);
solver.InsertEdge(P1(i + 3, j), P0(i, j), cost);
}
if (j + 3 < n) {
auto cost = GetCost(S, i, j + 1, i, j + 2);
solver.InsertEdge(P1(i, j), P0(i, j + 3), cost);
solver.InsertEdge(P1(i, j + 3), P0(i, j), cost);
}
}
}
}
auto result = solver.MaxFlow(source, sink);
if (result >= kInfinite) {
return -1;
}
return result;
} private:
int GetCost(const std::vector<std::string> &s, int x1, int y1, int x2,
int y2) {
if (s[x1][y1] == 'b' || s[x2][y2] == 'b') {
return kInfinite;
}
int ans = 0;
if (s[x1][y1] == '.') {
++ans;
}
if (s[x2][y2] == '.') {
++ans;
}
return ans;
}
};
code for problem3
constexpr int kMod = 1000000007;
constexpr int kMax = 2000; int f[kMax + 1][kMax + 1];
int C[kMax + 1][kMax + 1]; class Seatfriends {
public:
int countseatnumb(int N, int K, int G) {
f[1][1] = N;
for (int i = 1; i < K; ++i) {
for (int j = 1; j <= G; ++j) {
long long p = f[i][j];
if (p == 0) {
continue;
}
if (j < G) {
(f[i + 1][j + 1] += static_cast<int>(p * j % kMod)) %= kMod;
}
(f[i + 1][j - 1] += static_cast<int>(p * j % kMod)) %= kMod;
(f[i + 1][j] += static_cast<int>(p * 2 * j % kMod)) %= kMod;
}
}
if (K == N) {
return f[K][0];
} C[0][0] = 1;
for (int i = 1; i <= N; ++i) {
C[i][0] = C[i][i] = 1;
for (int j = 1; j < i; ++j)
C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % kMod;
} long long ans = 0;
for (int j = 1; j <= G; ++j) {
ans += static_cast<long long>(f[K][j]) * C[N - K - 1][j - 1] % kMod;
}
return static_cast<int>(ans % kMod);
}
};